= Solution
Write $u=\cos\theta$, $S=j/\kappa$ and use axial symmetry. The <radiative flux> is $F=2\pi\int_{-1}^1uI\,du$. Integrating the <radiative transfer equation> over directions gives
$$
\frac{dF}{d\tau}=4\pi(J-S).
$$
There are no energy sources in the atmosphere, so <radiative equilibrium> makes $F$ constant. Consequently
$$
\boxed{J=S=\frac{j}{\kappa}=\frac{\sigma T^4}{\pi}.}
$$
For the linear angular approximation $I=A+Cu$, the <radiation-field moments> are
$$
J=A,\qquad F=\frac{4\pi}{3}C,\qquad
cP_r=2\pi\int_{-1}^1u^2(A+Cu)\,du=\frac{4\pi}{3}A.
$$
Thus it satisfies the <Eddington closure approximation>, and
$$
\boxed{cP_r=\frac{4\pi}{3}J,\qquad C=\frac{3F}{4\pi}.}
$$
Since $F$ is constant, $C'=0$. Substitution in the <radiative transfer equation>, using $S=A$, gives $uA'=Cu$, hence $\boxed{A'=C}$.
In this moment approximation, the inward hemispheric flux at the surface is
$$
F_{\rm in}=-2\pi\int_{-1}^0u(A_0+Cu)\,du
=\pi A_0-\frac{2\pi}{3}C.
$$
Setting this boundary moment to zero gives $A_0=2C/3$. Therefore
$$
A(\tau)=\frac{3F}{4\pi}\left(\tau+\frac23\right).
$$
Combining it with the thermal <source function> and $F=\sigma T_e^4$ gives the <grey atmosphere> temperature law:
$$
\boxed{T^4=\frac34T_e^4\left(\tau+\frac23\right),\qquad
T_0=2^{-1/4}T_e.}
$$
\b[The surface condition is an approximate angular-moment condition.] If the affine intensity were interpreted as an exact nonnegative boundary radiation field, no incident radiation would require $I(0,u)=0$ for every $u<0$; a linear polynomial vanishing on that interval would have $A_0=C=0$ and no outward flux either. The derived formal profile indeed has $I(0,-1)=-F/(4\pi)<0$. This <positivity limitation of a linear Eddington intensity> is why one uses the <Eddington surface boundary condition> as a moment closure, not the affine ansatz as an exact emergent angular intensity. An outward direction-independent intensity with no inward intensity gives the same boundary moments.
For the gas-pressure-dominated atmosphere, combine $dP/dz=-\rho g$ with $d\tau/dz=-\kappa\rho$. Thus the <hydrostatic equilibrium in optical depth> is $dP/d\tau=g/\kappa$. Let $w=T^4$. The <grey atmosphere> law implies $dw/d\tau=3T_0^4/2$. With the specified power-law <opacity>,
$$
\frac{d(P^\alpha)}{dw}
=\frac{2\alpha g}{3\kappa_0T_0^4}\,w^{\beta-1}.
$$
At the outer vacuum limit take $P=0$ and $T=T_0$. Since $\alpha,\beta>0$, integration gives
$$
\boxed{P^\alpha=\frac{2\alpha g}{3\beta\kappa_0T_0^4}
\left(T^{4\beta}-T_0^{4\beta}\right).}
$$
This is the <grey-atmosphere pressure with power-law opacity> relation.
At the matching radius the local <radiative flux> is $L_r/(4\pi r^2)=\sigma T^4$. In the thin <plane-parallel atmosphere> approximation this equals $F=\sigma T_e^4$, so $T=T_e$ and $T^4=2T_0^4$. Multiplying the pressure relation by $\kappa_0T^{4-4\beta}/g$ yields the required \b[surface pressure-opacity relation]
$$
\boxed{\frac{P\kappa}{g}
=\frac{2\alpha}{3\beta}\frac{T^4}{T_0^4}
\left[1-\left(\frac{T_0}{T}\right)^{4\beta}\right]
=\frac{4\alpha}{3\beta}(1-2^{-\beta}).}
$$
Here $P\kappa$ is a product, not a pressure subscript.
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