= Solution
The stellar luminosity is $L_\star=4\pi R_\star^2\sigma_{\rm SB}T_\star^4$. At separation $a$, the planet intercepts a disk of area $\pi R_p^2$. For <Bond albedo> $A_B$ and uniform emission over its full surface, energy balance gives
$$
(1-A_B)\frac{L_\star}{4\pi a^2}\pi R_p^2=4\pi R_p^2\sigma_{\rm SB}T_e^4,
\qquad
\boxed{T_e=T_\star\left(\frac{R_\star}{2a}\right)^{1/2}(1-A_B)^{1/4}.}
$$
This <planetary equilibrium temperature> neglects intrinsic heat. Dayside-only or nonuniform emission changes the redistribution factor.
For the <single-layer greenhouse model>, assume a black infrared surface, an atmosphere transparent to incoming stellar radiation, and infrared absorptivity/emissivity $\alpha$. Let $T_s$ be surface temperature and $T_a$ layer temperature. Kirchhoff's law makes the layer's emission in each direction $\alpha\sigma_{\rm SB}T_a^4$. Its balance is
$$
\alpha\sigma_{\rm SB}T_s^4=2\alpha\sigma_{\rm SB}T_a^4,
$$
so $T_a^4=T_s^4/2$ for $\alpha>0$. Surface balance includes the downward layer emission:
$$
\sigma_{\rm SB}T_e^4+\alpha\sigma_{\rm SB}T_a^4=\sigma_{\rm SB}T_s^4.
$$
Therefore
$$
\boxed{T_s=T_e(1-\alpha/2)^{-1/4}.}
$$
The no-atmosphere limit is $T_s=T_e$, and a completely IR-absorbing single layer gives \b[$T_s=2^{1/4}T_e$]. The outgoing flux $(1-\alpha)\sigma T_s^4+\alpha\sigma T_a^4$ remains $\sigma T_e^4$. The layer delays escape by absorption and re-emission; it does not permanently retain a fraction $\alpha$ of every emitted photon, which would incorrectly predict a divergent temperature at $\alpha=1$.
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