= Solution
Use the printed <stellar gas-pressure fraction>, $\beta=P_g/P$, with $P=P_g+P_{\rm rad}$, and keep composition fixed. For a monatomic <perfect gas> plus equilibrium <blackbody radiation>, the <specific heats of a monatomic gas-radiation mixture> follow from its <specific internal energy> and <specific enthalpy>:
$$
u=\frac32\mathcal RT+\frac{aT^4}{\rho},\qquad h=u+\frac P\rho=\frac52\mathcal RT+\frac{4aT^4}{3\rho}.
$$
At fixed total <pressure>, differentiating the equation of state gives
$$
0=\beta\,d\log\rho+(4-3\beta)\,d\log T,\qquad\left(\frac{\partial\log\rho}{\partial\log T}\right)_P=-\frac{4-3\beta}{\beta}.
$$
In particular, $\beta$ is not held constant during that derivative. The <specific heat capacity at constant pressure> is $c_P=(\partial h/\partial T)_P$, so
$$
c_P=\frac52\mathcal R+\frac{4aT^3}{3\rho}\left(4+\frac{4-3\beta}{\beta}\right).
$$
Since $aT^3/(3\rho)=\mathcal R(1-\beta)/\beta$, this simplifies to
$$
\boxed{c_P=\frac{\mathcal R}{2\beta^2}(32-24\beta-3\beta^2)=\frac{k}{2\mu H\beta^2}(32-24\beta-3\beta^2).}
$$
The pure-gas limit is $5\mathcal R/2$. For a gas with unspecified molecular degrees of freedom, replace $3\mathcal R/2$ in $u$ by its gas heat capacity $c_{V,g}$: then $c_P=c_{V,g}+\mathcal R+4\mathcal R(1-\beta)(4+\beta)/\beta^2$. The boxed formula uses the conventional monatomic stellar-gas interpretation.
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