= Solution
Use downward <optical depth> $t$ measured from the top, with total $D=\tau_\infty$. This is the effective optical-depth/<two-stream approximation>[two-stream] convention that gives an attenuation factor $e^{-t}$. Constant absorption per unit <mass> and constant gravity give $t=Dp/p_s$. Write the locally emitted flux source as $S(t)=\pi B(t)=S_0+bt$.
The atmospheric contribution to upward outgoing longwave flux is
$$
F_{\rm atm}=\int_0^D S(t)e^{-t}\,dt
=S_0(1-e^{-D})+b[1-(D+1)e^{-D}].
$$
Because the source is linear, the <effective emission pressure> makes this exactly $(1-e^{-D})S(t_{\rm eff})$, where
$$
\boxed{t_{\rm eff}=\frac{1-(D+1)e^{-D}}{1-e^{-D}}
=1-\frac{D}{e^D-1},\qquad
p_{\rm rad}=\frac{p_s}{D}t_{\rm eff}.}
$$
Thus for $D\gg1$, $t_{\rm eff}\to1$ and $p_{\rm rad}\simeq p_s/D$. For $D\ll1$, expansion gives $t_{\rm eff}=D/2-D^2/12+O(D^4)$, so $p_{\rm rad}\simeq p_s/2$. In the thin limit $F_{\rm atm}\simeq D\,\pi B(p_s/2)$: it is the small atmospheric contribution, not the whole outgoing flux, which this level characterizes. Surface radiation transmitted through the column must be added separately.
\b[Angular convention:] these numerical coefficients use an effective <optical depth>. If $t$ instead means literal vertical monochromatic <optical depth> and the angular <integral> is retained exactly, the flux weight is $2E_2(t)$, not $e^{-t}$. Its thick-limit mean depth is $2/3$, giving $p_{\rm rad}\simeq2p_s/(3D)$ for a linear source. The coefficient one in the requested expression therefore specifies a <two-stream approximation>[two-stream/diffusivity] convention; it is not an angle-independent identity. The thin-limit half-column result survives to leading order.
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