= Solution
For $a\gg a_{AB}$, both stars illuminate the <circumbinary planet> at approximately distance $a$. Treat them as <blackbodies>, neglect internal heating, and assume <Bond albedo> $A_B$ and uniform global reradiation. The absorbed and emitted powers are
$$
\pi R_p^2(1-A_B)\sigma_{\rm SB}\frac{R_A^2T_A^4+R_B^2T_B^4}{a^2}
=4\pi R_p^2\sigma_{\rm SB}T_p^4.
$$
Thus the <equilibrium temperature of a circumbinary planet> is
$$
\boxed{T_p=\left[\frac{1-A_B}{4a^2}(R_A^2T_A^4+R_B^2T_B^4)\right]^{1/4}.}
$$
For an example, take $R_A=R_\odot$, $R_B=0.3R_\odot$, $T_A=6000\,\mathrm K$, $T_B=3000\,\mathrm K$, $a=1\,\mathrm{au}$, $a_{AB}=0.05\,\mathrm{au}$, circular coplanar orbits, and $A_B=0$. Then
$$
\boxed{T_p\simeq290\,\mathrm K.}
$$
The cooler star contributes only $0.3^2(3000/6000)^4\simeq0.0056$ of the hotter star's luminosity in this example.
An opaque isothermal atmosphere emits $F_\lambda=\pi B_\lambda(T_p)$, with no absorption or emission bands. The <Wien displacement law> puts the maximum of this wavelength spectrum near $\lambda_{\rm max}\simeq10\,\mu\mathrm m$. At observer distance $d$, the planetary spectral flux is $\pi B_\lambda(T_p)(R_p/d)^2$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-315-isothermal-spectrum.png]
{title=Isothermal circumbinary-planet emission spectrum}
{description=The emergent surface flux for the example above. The wavelength spectrum peaks near ten micrometres; the exactly isothermal opaque model has no molecular bands.}
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