= Solution
Write $u=\cos\theta$ for the direction cosine and $B=j/\kappa=\sigma T^4/\pi$ for the <radiative transfer source function>. Integrating the <radiative transfer equation> $u\,\partial_\tau I=I-B$ over <solid angle> gives
$$
\frac{dF}{d\tau}=\int(I-B)\,d\Omega=4\pi(J-B).
$$
<Radiative equilibrium> makes the <radiative flux> $F$ constant, so \b[$J=B=j/\kappa$].
For $I=A+Cu$, the <radiation-field moments> are
$$
J=\frac12\int_{-1}^1I\,du=A,\qquad F=2\pi\int_{-1}^1uI\,du=\frac{4\pi C}{3},\qquad cP_r=2\pi\int_{-1}^1u^2I\,du=\frac{4\pi A}{3}.
$$
Consequently the <Eddington closure approximation> holds. Substituting into the <radiative transfer equation>, with $B=J=A$, gives $uA'+u^2C'=Cu$. Equality for every $u$ requires $A'=C$ and $C'=0$, with $C=3F/(4\pi)$. The stipulated inward hemispheric <radiative flux> at the surface is
$$
F_{\rm in}=2\pi\int_{-1}^0u(A_0+Cu)\,du=-\pi A_0+\frac{2\pi C}{3}.
$$
Putting this equal to zero fixes $A_0=2C/3$, hence $A=C(\tau+2/3)$. Using $A=\sigma T^4/\pi$ and the <effective temperature> definition $F=\sigma T_e^4$ gives the <grey atmosphere> law
$$
\boxed{T^4=\frac34T_e^4\left(\tau+\frac23\right)},\qquad \boxed{T_0=2^{-1/4}T_e}.
$$
The <positivity limitation of a linear Eddington intensity> matters here: the formal surface intensity $I(0,u)=C(2/3+u)$ is negative for $u<-2/3$. Thus this angular ansatz gives the requested <Eddington surface boundary condition> for approximate moments; vanishing integrated inward flux is not an exact, nonnegative, pointwise no-incoming radiation field.
Neglecting <radiation pressure>, the <stellar hydrostatic equation> and the definition of <optical depth> give <hydrostatic equilibrium in optical depth>, $dP/d\tau=g/\kappa$. Since $T^4=T_0^4(1+3\tau/2)$, putting $v=T^4$ gives
$$
\frac{d(P^\alpha)}{dv}=\frac{2\alpha g}{3\kappa_0T_0^4}v^{\beta-1}.
$$
Taking $P(0)=0$ and $\alpha>0$, the <grey-atmosphere pressure with power-law opacity> is therefore
$$
\boxed{P^\alpha=\frac{2\alpha g}{3\beta\kappa_0T_0^4}\left(T^{4\beta}-T_0^{4\beta}\right)}\qquad(\beta\ne0).
$$
\b[The pressure formula printed in the PDF is missing the factor $1/\beta$.] Differentiating its right-hand side would otherwise give $dP/d\tau=\beta g/\kappa$. For $\beta=0$ the correct <limit> is $P^\alpha=2\alpha g\log(T^4/T_0^4)/(3\kappa_0T_0^4)$.
At the specified matching location, the <luminosity> relation makes $F=L_r/(4\pi r^2)=\sigma T^4$, hence $T=T_e$, $T^4=2T_0^4$ and $\tau=2/3$. Multiplying the corrected pressure relation by $\kappa_0T^{4-4\beta}/g$ gives
$$
\boxed{\frac{P\kappa}{g}=\frac{4\alpha}{3\beta}\left(1-2^{-\beta}\right)}.
$$
This agrees with the PDF's final <surface boundary condition for a power-law-opacity grey atmosphere>. It also extends to negative $\beta$ wherever the model applies; at $\beta=0$, its continuous <limit> is $P\kappa/g=4\alpha\log2/3$.
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