= Solution
Use the <Fourier transform> convention $\widehat\varphi(\xi)=\int e^{-ix\cdot\xi}\varphi(x)\,dx$ and $D_j=-i\partial_j$, so $\widehat{P(D)u}=P(\xi)\widehat u$. All <distribution> pairings use complex <bilinearity>, and the <formal transpose> is $P(-D)$, without <complex conjugation> of the coefficients. If $D_j$ instead denotes $\partial_j$, the Fourier multiplier is $P(i\xi)$; the argument below is unchanged after this substitution.
For a nonempty <compact convex set> $K\subset\mathbb R^n$, let $H_K(\eta)=\sup_{x\in K}x\cdot\eta$ be its <support function>. The <Paley–Wiener–Schwartz theorem> says that the <Fourier transform> is a bijection between <distributions> supported in $K$ and <entire functions> $F$ on $\mathbb C^n$ for which
$$
\boxed{|F(\zeta)|\leq C(1+|\zeta|)^N e^{H_K(\operatorname{Im}\zeta)}}
$$
for some $C>0$ and nonnegative integer $N$. In particular, support in $\overline B_R$ is equivalent to the bound with $e^{R|\operatorname{Im}\zeta|}$. Convexity is essential in the precise support formulation: the <support function> of a set equals that of its <convex hull>.
Suppose first that $u$ is a <compactly supported distribution> with support in $K$. Define $F(\zeta)=\langle u,e^{-ix\cdot\zeta}\rangle$, inserting any <cutoff function> equal to one near $K$ when viewing $u$ on <test functions>. A fixed cutoff and the finite <order of a distribution> estimate permit differentiation of the pairing in each complex variable, giving $\partial_\zeta^\alpha F=\langle u,(-ix)^\alpha e^{-ix\cdot\zeta}\rangle$. The <exponential function>'s <power series> converges with every required derivative on a fixed compact set; hence $F$ is an <entire function>. On real frequencies it agrees with the <Fourier transform of a compactly supported distribution>.
The exact exponential indicator needs a shrinking <cutoff function>, rather than one fixed outside $K$. Choose $\chi_\varepsilon=1$ near $K$, with support in $K+B_{2\varepsilon}$ and $\|\partial^\alpha\chi_\varepsilon\|_\infty\leq C_\alpha\varepsilon^{-|\alpha|}$ for $0<\varepsilon\leq1$. Such a cutoff is obtained by convolving the indicator of $K+B_\varepsilon$ with a <mollifier> supported in $B_{\varepsilon/2}$. On one fixed ball containing all these supports, the finite <order of a distribution> bound is
$$
|\langle u,\theta\rangle|\leq C_0\max_{|\alpha|\leq m}\|\partial^\alpha\theta\|_\infty.
$$
The <Leibniz rule> applied to $\theta=\chi_\varepsilon e^{-ix\cdot\zeta}$ therefore gives
$$
|F(\zeta)|\leq C_1\varepsilon^{-m}(1+|\zeta|)^m
e^{H_K(\operatorname{Im}\zeta)+2\varepsilon|\operatorname{Im}\zeta|}.
$$
Taking $\varepsilon=(1+|\zeta|)^{-1}$ proves the required bound, with $N=2m$ and a fixed additional factor at most $e^2$.
Conversely, suppose $F$ is an <entire function> with the displayed bound. Its real restriction has <polynomial growth>, so its inverse <Fourier transform> defines a <tempered distribution> by
$$
\langle u,\varphi\rangle=(2\pi)^{-n}\int_{\mathbb R^n}F(\xi)\widehat\varphi(-\xi)\,d\xi,
\qquad \varphi\in\mathcal S.
$$
Rapid decay of the <Schwartz function> $\widehat\varphi$ makes this integral absolutely convergent and continuous in the <Schwartz space> topology.
To determine the <support of a distribution>, take a <test function> supported in a half-space $x\cdot\omega\geq H_K(\omega)+\delta$, where $|\omega|=1$ and $\delta>0$. The <contour-shift proof of the Paley–Wiener–Schwartz theorem> gives, for each $t\geq0$,
$$
\langle u,\varphi\rangle=(2\pi)^{-n}\int_{\mathbb R^n}
F(\xi+it\omega)\widehat\varphi(-\xi-it\omega)\,d\xi.
$$
Here is the estimate that justifies both the shift and its limiting use. Since $\widehat\varphi(-\xi-it\omega)=\int e^{ix\cdot\xi}e^{-t x\cdot\omega}\varphi(x)\,dx$, repeated <integration by parts> with $1-\Delta_x$ yields
$$
|\widehat\varphi(-\xi-it\omega)|
\leq C_L(1+t)^{2L}e^{-t(H_K(\omega)+\delta)}(1+|\xi|^2)^{-L}.
$$
For fixed $t$, the same estimate is uniform over the imaginary strip from $0$ to $t\omega$. Rotate coordinates so $\omega$ is the first coordinate direction, apply the <Cauchy integral theorem> on truncated rectangles, and let their real edges tend to infinity. If $2L>N+n$, the boundary integrals vanish and the horizontal integrals converge absolutely; thus the claimed shift is valid. Using $H_K(t\omega)=tH_K(\omega)$ now bounds the pairing by
$$
C'(1+t)^{N+2L}e^{-\delta t}
\int_{\mathbb R^n}(1+|\xi|)^{N-2L}\,d\xi,
$$
which tends to zero. A <separating hyperplane> exists at every point outside the <compact convex set> $K$. A finite <partition of unity> on the support of any exterior <test function> reduces it to such half-spaces, so $\operatorname{supp}u\subseteq K$. <Fourier inversion> gives uniqueness; the entire transform of this $u$ agrees with $F$ on $\mathbb R^n$ and hence everywhere, by applying the one-variable <identity theorem> successively in each coordinate. This completes both directions.
Now apply this theorem to <compact support solvability for a constant-coefficient ordinary differential equation>. If $P$ is a nonzero constant, its kernel is zero and $u=v/P$ is the unique solution. Otherwise write
$$
P(\zeta)=a_m\prod_{j=1}^m(\zeta-\alpha_j),\qquad \alpha_j\ne\alpha_k\quad(j\ne k).
$$
The solution space of $P(-D)\varphi=0$ is
$$
\mathcal N=\operatorname{span}\{e^{-i\alpha_1x},\ldots,e^{-i\alpha_mx}\}.
$$
Indeed these exponentials solve the equation, their initial derivative vectors have a nonzero <Vandermonde determinant>, and uniqueness for the order-$m$ <ordinary differential equation> makes them a basis. Pairing $P(D)u=v$ with any such solution proves necessity:
$$
\langle v,\varphi\rangle=\langle u,P(-D)\varphi\rangle=0.
$$
For sufficiency, choose $R$ with $\operatorname{supp}v\subseteq[-R,R]$ and let $V(\zeta)=\langle v,e^{-ix\zeta}\rangle$. The assumed annihilation is precisely $V(\alpha_j)=0$ for every $j$. Since all <roots of a polynomial> are simple, $U=V/P$ has removable singularities at every root and is an <entire function>. Outside one disk, $|P(\zeta)|\geq c(1+|\zeta|)^m$. Inside that disk, the extended $U$ is bounded. The <polynomial division preservation of exponential type> therefore gives
$$
|U(\zeta)|\leq C'(1+|\zeta|)^{N'}e^{R|\operatorname{Im}\zeta|}.
$$
The <Paley–Wiener–Schwartz theorem> produces a <compactly supported distribution> $u$ with transform $U$, supported in the same interval. Then $\widehat{P(D)u}=PU=V$, so <Fourier inversion> gives $P(D)u=v$. Thus
$$
\boxed{v\in P(D)\mathcal E'(\mathbb R)\ \Longleftrightarrow\
\langle v,\varphi\rangle=0\ \text{for every }\varphi\in\mathcal N.}
$$
The compactly supported solution is unique: $P\widehat u=0$ forces the entire transform to vanish off the finitely many roots and hence everywhere. With repeated roots, the corresponding condition would also require derivatives $V^{(k)}(\alpha)=0$, or pairing with $x^k e^{-i\alpha x}$, up to one less than each multiplicity.
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