Surjectivity of differential operators with constant coefficients

Here is another approach. Let $R$ be a non-zero homogeneous polynomial of degree $n$. We want to show that the mapping $Q\mapsto R(\partial)Q$ is surgective from $V_{m+n}$ to $V_m$ where $V_k$ is the space of homogeneous polynomials of degree $k$. Note that $\langle A,B\rangle=[A(\partial)\bar B](0)$ is a scalar product on $V_k$ (with monomials forming an orthogonal basis). If our mapping is not surjective, then there exists a non-zero polynomial $S\in V_m$ such that $\langle S, R(\partial) Q\rangle=\langle S\bar R,Q\rangle=0$ for all $Q\in V_{n+m}$. But $S\bar R$ is a non-zero polynomial, so taking $Q=S\bar R$, we get a contradiction.

Too long for a comment. I want to use a version of the Lojaciewicz theorem of division of distributions by an analytic function (in fact Hörmander's result of division by a polynomial). We may assume that $P(x) =x^\alpha=x_1^{\alpha_1}\dots x_n^{\alpha_n}$ a monomial homogeneous with degree $\vert \alpha\vert=\alpha_1+\dots +\alpha_n$. Let us replace your notation $D$ by an operator $$A(D)=\sum_{\vert \beta\vert =m}a_\beta D^\beta,\quad D=-i\nabla.$$ The question at hand is to find an homogeneous polynomial $Q$ such that $ A(D) Q= x^\alpha. $ By Fourier transformation, it is equivalent to solve $$ A(\xi)\widehat Q(\xi)=i^{\vert \alpha\vert}\delta^{(\alpha)}, $$ which means divide a derivative of the Dirac mass by an homogeneous ($A$ is assumed to be non-zero) polynomial. It is indeed possible by the aforementioned results of division and we find that $\widehat Q$ is homogeneous with degree $-n-\vert \alpha\vert-m$, so that $Q$ is homogeneous with degree $\vert \alpha\vert+m$. Moreover $\widehat Q$ is supported at the origin, proving that $Q$ is a polynomial.