= Solution
In the thin <radiative envelope>, take enclosed mass $m(r)\simeq M$, luminosity $L_r\simeq L$, and constant <mean molecular weight> $\mu$. Combining <radiative diffusion in a star> with the <hydrostatic pressure support equation> gives
$$
\nabla_{\rm rad}=\frac{3\kappa PL}{16\pi a_{\rm rad}cGMT^4}.
$$
The <ideal gas> relation is $\rho=AP/T$, where $A=\mu m_u/k_B$, so the opacity law implies
$$
\boxed{a=n+1,\qquad b=n+s+4,\qquad C=\frac{16\pi a_{\rm rad}cGM}{3\kappa_0A^nL},\qquad \nabla=\frac1C\frac{P^a}{T^b}.}
$$
Here $a_{\rm rad}$ is the <radiation constant>, distinct from the exponent $a$. Since $\nabla=(P/T)dT/dP$, separation gives $P^{a-1}dP=CT^{b-1}dT$. With photospheric $T=T_{\rm eff}$, the <power-law opacity radiative envelope> therefore has
$$
\boxed{P(T)=\left[P_{\rm phot}^{\,a}+\frac{aC}{b}\left(T^b-T_{\rm eff}^{\,b}\right)\right]^{1/a}.}
$$
This expression assumes $a,b\ne0$ and a positive bracket on the physical branch. If $b=0$, replace $(T^b-T_{\rm eff}^b)/b$ by $\log(T/T_{\rm eff})$; if $a=0$, the integrated relation is $\log(P/P_{\rm phot})=C(T^b-T_{\rm eff}^b)/b$, with the same logarithmic limit when $b=0$.
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