= Solution
Assume a circular orbit of radius $a$, a spherical planet, bolometric <Bond albedo> $A_B$, negligible internal heat, unit thermal emissivity and complete redistribution of absorbed heat over the sphere. The stellar <luminosity> is $L_s=4\pi R_s^2\sigma T_s^4$, and the planet absorbs the incident <radiative flux> through its projected cross-section:
$$
P_{\rm abs}=\pi R_p^2(1-A_B)\frac{L_s}{4\pi a^2}.
$$
Thermal reradiation is $P_{\rm emit}=4\pi R_p^2\sigma T_{\rm eq}^4$. Equating these gives the <planetary equilibrium temperature>
$$
\boxed{T_{\rm eq}=\left[\frac{(1-A_B)L_s}{16\pi\sigma a^2}\right]^{1/4}=T_s\sqrt{\frac{R_s}{2a}}(1-A_B)^{1/4}.}
$$
The planet's radius cancels. A non-black thermal emissivity $\epsilon$ divides the absorbed flux by $\epsilon$ in the fourth-power balance. For uniform dayside-only reradiation, the emitting area is $2\pi R_p^2$ and $T_{\rm day}=2^{1/4}T_{\rm eq}$. With no local redistribution, the substellar point has $T_{\rm sub}=\sqrt2T_{\rm eq}$, while other points depend on incidence angle. These are different temperature conventions, not contradictory formulas.
The <planetary equilibrium temperature> is not the surface greenhouse temperature or the <internal effective temperature of a planet>. If intrinsic cooling matters and both powers escape through the same emitting area, the total <effective temperature> obeys $T_{\rm eff}^4=T_{\rm int}^4+T_{\rm eq}^4$.
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