= Solution
Take <parity symmetry in quantum field theory> to preserve free time evolution, so the conjugate initial data reflect along with the field. Substitute the <mode expansion of a free field> into the defining <parity operator> relation and change the integration variable $\mathbf p\mapsto-\mathbf p$. Since $E_{-\mathbf p}=E_{\mathbf p}$, coefficient comparison, or comparison of both the field and its conjugate initial data, gives the intrinsic-even convention
$$
\boxed{\begin{aligned}
\mathcal P a_{\mathbf p}\mathcal P^\dagger&=a_{-\mathbf p},&
\mathcal P b_{\mathbf p}\mathcal P^\dagger&=b_{-\mathbf p},\\
\mathcal P a_{\mathbf p}^\dagger\mathcal P^\dagger&=a_{-\mathbf p}^\dagger,&
\mathcal P b_{\mathbf p}^\dagger\mathcal P^\dagger&=b_{-\mathbf p}^\dagger.
\end{aligned}}
$$
Here <parity symmetry in quantum field theory> reverses spatial <momentum> and preserves time. The <creation operator> formulas follow by taking <Hermitian adjoints>. The given field relation has fixed the intrinsic phase to $+1$.
For one <real scalar field>, let $R$ act on one-particle <wave functions> by $(Rf)(\mathbf p)=f(-\mathbf p)$. It is a <Hermitian operator> with $R^2=1$, so $P_-=(1-R)/2$ is the <orthogonal projector> onto odd <wave functions>. Its <second quantization> is the odd-mode <number operator>
$$
N_{\rm odd}=\frac12\int_{\mathbf p}a_{\mathbf p}^\dagger(a_{\mathbf p}-a_{-\mathbf p})
=\frac14\int_{\mathbf p}(a_{\mathbf p}^\dagger-a_{-\mathbf p}^\dagger)(a_{\mathbf p}-a_{-\mathbf p}).
$$
The full <momentum space> is integrated in both expressions; the factors prevent double counting the pairs $\{\mathbf p,-\mathbf p\}$. A suitable <parity operator> is
$$
\boxed{\mathcal P=\exp(i\pi N_{\rm odd})
=\exp\left[\frac{i\pi}{2}\int_{\mathbf p}a_{\mathbf p}^\dagger(a_{\mathbf p}-a_{-\mathbf p})\right].}
$$
For verification, $[N_{\rm odd},a_{\mathbf p}]=-(a_{\mathbf p}-a_{-\mathbf p})/2$. Thus the even combination $a_{\mathbf p}+a_{-\mathbf p}$ is unchanged, and the odd combination $a_{\mathbf p}-a_{-\mathbf p}$ changes sign under conjugation by $e^{i\pi N_{\rm odd}}$, exactly reproducing $a_{\mathbf p}\mapsto a_{-\mathbf p}$. The same holds for <creation operators>. The odd-mode <number operator> has integer <eigenvalues>, so $\mathcal P^2=1$, and $\mathcal P|0\rangle=|0\rangle$. Multiplying by an overall constant phase would not change its conjugation action. This is <parity as odd-mode number parity>.
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