= Solution
For $z\ne0$, the <coefficient> symmetry yields
$$
z^{2d}\overline{P(1/\overline z)}
=z^{2d}\overline{(1/\overline z)^dp(1/\overline z)}
=z^dp(z)=P(z).
$$
Thus $P$ equals its reversed-conjugate <polynomial>. If $\zeta\ne0$ is a <root of a polynomial>, with $P(\zeta)=0$, this identity gives $P(1/\overline\zeta)=0$, proving
$$
\boxed{\zeta\text{ a root}\ \Longrightarrow\ 1/\overline\zeta\text{ a root}}.
$$
The <reciprocal-conjugate root pairing> also preserves the <multiplicity of a root>: reversal and <complex conjugation> take each factor associated with $\zeta$ to the corresponding factor associated with $1/\overline\zeta$, with the same exponent.
Since $P(0)=p_{-d}\ne0$, no zero <root of a polynomial> occurs, so the reciprocal operation is always defined on the <roots of a polynomial>. Roots off the <unit circle> are paired on opposite sides of it; <roots of a polynomial> on the <unit circle> are fixed by this operation. Pairing alone does not force even multiplicity at those fixed <roots of a polynomial>; that extra fact uses nonnegativity.
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