= Solution
Divide the radiation-pressure gradient by the total-pressure gradient. <Radiative diffusion in a star> and <hydrostatic equilibrium> give
$$
\frac{dP_r}{dr}=-\frac{\kappa\rho L_r}{4\pi c r^2},\qquad
\frac{dP}{dr}=-\frac{Gm_r\rho}{r^2},\qquad
\frac{dP_r}{dP}=\frac{\kappa L_r}{4\pi cGm_r}
=\frac{\alpha L}{4\pi cGM}=:f.
$$
This last quantity is independent of radius under the specified <opacity> assumption. Integration gives $P_r=fP+C$. With the standard idealized zero-pressure outer boundary, where both components vanish, $C=0$. The <stellar gas-pressure fraction> is therefore
$$
\boxed{\beta=\frac{P_g}{P}=1-f
=1-\frac{\alpha L}{4\pi cGM},}
$$
\b[constant throughout the model]. This is the <Eddington standard model> idealization. A finite photospheric <pressure> offset would need its boundary treatment; the differential relation alone does not set the integration constant to zero.
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