= Solution
Repeated <integration by parts> against a compactly supported <test function> gives the bilinear <formal adjoint>
$$
\boxed{P^*v=\sum_{|\alpha|\le m}(-1)^{|\alpha|}\partial^\alpha(a_\alpha v).}
$$
There is no boundary remainder because $v$ has <compact support>. The product rule makes this an order-at-most-$m$ linear differential operator with <real analytic> coefficients. The displayed integral pairing has no complex conjugation, so this is the formal transpose; for a sesquilinear pairing the coefficients would be conjugated.
For uniqueness, suppose $Q$ is the difference of two candidates. Then $\int uQv=0$ for every smooth compactly supported $u$ and $v$, so $Qv=0$ for every such $v$. At an interior point, a <polynomial> times a cutoff can prescribe every <derivative> of $v$ through order $m$ independently. Hence each coefficient of $Q$ is zero. This proves uniqueness as a differential operator, not only uniqueness of its integrated action.
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