= Solution
Apply the <parity symmetry in quantum field theory> to the given <mode expansion of a Dirac field>:
$$
\widehat P\psi(x)\widehat P^{-1}
=\sum_{p,s}\left[\eta_Pb^s(p_P)u^s(p)e^{-ip\cdot x}
-\eta_Pd^{s\dagger}(p_P)v^s(p)e^{ip\cdot x}\right].
$$
The stated <Dirac spinor> identities are equivalently $u^s(p)=\gamma^0u^s(p_P)$ and $v^s(p)=-\gamma^0v^s(p_P)$. Relabel the momentum sum by $p\mapsto p_P$ and use $p_P\cdot x=p\cdot x_P$ to obtain
$$
\widehat P\psi(x)\widehat P^{-1}
=\eta_P\gamma^0\sum_{p,s}\left[b^s(p)u^s(p)e^{-ip\cdot x_P}+d^{s\dagger}(p)v^s(p)e^{ip\cdot x_P}\right]
=\boxed{\eta_P\gamma^0\psi(x_P)}.
$$
The phase $|\eta_P|=1$ is the intrinsic parity convention.
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