= Solution
An adiabatic change of box size preserves the mode labels, whose momenta scale as $p\propto L^{-1}$. Since $V=L^3$, differentiation gives $dp/dV=-p/(3V)$. <Occupation numbers> are fixed under this slow mechanical change, so summing the changes in the individual particle energies gives
$$
dE=\int_0^\infty E'(p)\,\overline n(p)\,dp\left(-\frac{p}{3V}dV\right).
$$
Comparison with $dE=-P\,dV$ at fixed <entropy> and particle number yields
$$
\boxed{P=\frac1{3V}\int_0^\infty pE'(p)\overline n(p)\,dp.}
$$
The integration notation means summing over the original momentum-mode labels; it does not assume that the density in momentum space is unchanged after rescaling. In the ultrarelativistic limit $E(p)=cp$, so $pE'(p)=E(p)$ and \b[$P=E/(3V)=\rho c^2/3$].
Adiabatic radiation obeys $d(Ea^3)=-P\,d(a^3)$, giving <energy density> proportional to $a^{-4}$. Equilibrium <photon energy density> is proportional to $T^4$, so $T\propto a^{-1}$. Therefore the photon number in a <comoving volume>, $n_\gamma a^3\propto T^3a^3$, is constant. Its <entropy> is also constant: photon <entropy density> is $(\epsilon_\gamma+P)/T\propto T^3$. Photon number is not a microscopic <conserved charge>, but equilibrium processes maintaining the spectrum do not alter this comoving-number scaling in <adiabatic expansion>; entropy-producing processes would invalidate the assumptions.
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