= Solution
For $z$ on the <unit circle>, use the conjugated monomial <vector>
$$
w(z)=(1,z^{-1},\ldots,z^{-d})^T
=(1,\overline z,\ldots,\overline z^{\,d})^T.
$$
Then the prescribed diagonal sums give
$$
w(z)^*Mw(z)=\sum_{i,j=0}^dM_{ij}z^{i-j}
=\sum_{k=-d}^d p_kz^k=p(z).
$$
<Positive semidefiniteness> proves
$$
\boxed{p(z)=w(z)^*Mw(z)\geq0\quad(|z|=1)}.
$$
This is the <Gram matrix representation of a trigonometric polynomial>. The Hermitian condition also implies $p_{-k}=\overline{p_k}$, so its values on the <unit circle> are real. The conjugated monomial <vector> is required by the source's $i-j=k$ convention; the unconjugated <vector> would represent $p(z^{-1})$ instead.
In particular $p_0=\operatorname{tr}M\geq0$. If this <matrix trace> is zero, all nonnegative <eigenvalues> vanish and $M=0$, so $p=0$. A general feasible <Gram matrix> need not have <matrix rank> one; the <Fejér–Riesz theorem> ensures a rank-one representative exists whenever the nonnegative <trigonometric polynomial> is nonzero.
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