= Solution
Every positive-energy <Dirac spinor> obeys $(\not p-m)u^s(\mathbf p)=0$. Therefore the <fermion spin sum> satisfies
$$
(\not p-m)P=\sum_s[(\not p-m)u^s]\bar u^s=0.
$$
Since $\bar u^s=u^{s\dagger}\gamma^0$ and $(\gamma^0)^2=I$, the <trace> of each outer product is its contracted scalar. The prescribed normalization and the two <spin> states yield
$$
\boxed{\operatorname{tr}(P\gamma^0)=\sum_su^{s\dagger}u^s=2(2E_p)=4E_p.}
$$
Now make exactly the stated <polynomial> assumption and apply <polynomial reduction of a slashed vector>. By <2/a>, on the <mass shell> $p^2=m^2$ write $P=A\not p+BI$. Then
$$
0=(\not p-m)P=(Am^2-mB)I+(B-mA)\not p.
$$
The <matrices> $I$ and $\not p$ are linearly independent for a physical nonzero $p$, so $B=mA$. The <gamma matrix trace identities>, $\operatorname{tr}\gamma^0=0$ and $\operatorname{tr}(\gamma^\mu\gamma^0)=4\eta^{\mu0}$, give
$$
4E_p=\operatorname{tr}[A(\not p+m)\gamma^0]=4AE_p.
$$
Thus $A=1$ and
$$
\boxed{P(\mathbf p)=\not p+m.}
$$
No explicit representation of the <Dirac spinors> has been used. The same reasoning covers $m=0$ provided $p^0>0$.
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