= Solution
Convection begins when the <radiative temperature gradient> equals the <adiabatic temperature gradient> $\nabla_{\rm ad}$. To convert optical depth into pressure, assume a pressure-law opacity $\kappa=\kappa_0P^n$ and constant gravity. Hydrostatic balance gives
$$
\tau(P)=\frac{\kappa_0P^{n+1}}{(n+1)g},
\qquad
\frac{d\log\tau}{d\log P}=n+1.
$$
Hence
$$
\nabla_{\rm rad}
=\frac{d\log T}{d\log P}
=\frac{(n+1)\tau[B-\beta Ce^{-\beta\tau}]}
{4[A+B\tau+Ce^{-\beta\tau}]}.
$$
The exact <radiative-convective boundary> is the positive solution of
$$
\boxed{
\frac{(n+1)\tau_{\rm rc}[B-\beta Ce^{-\beta\tau_{\rm rc}}]}
{4[A+B\tau_{\rm rc}+Ce^{-\beta\tau_{\rm rc}}]}
=\nabla_{\rm ad},
\qquad
P_{\rm rc}=\left[
\frac{(n+1)g\tau_{\rm rc}}{\kappa_0}
\right]^{1/(n+1)}}.
$$
Deep enough that the exponential term is negligible,
$$
\tau_{\rm rc}\simeq
\frac{4\nabla_{\rm ad}A}
{B(n+1-4\nabla_{\rm ad})},
$$
which requires $n+1>4\nabla_{\rm ad}$. This exposes why constant opacity is inadequate for a molecular atmosphere: its limiting radiative gradient is $1/4$, below $\nabla_{\rm ad}\simeq2/7$.
In a grey scaling, $A\sim T_{\rm eq}^4$ measures irradiation and $B\sim T_{\rm int}^4$ measures intrinsic flux. Thus
$$
P_{\rm rc}\propto(A/B)^{1/(n+1)}.
$$
For a hot Jupiter with $T_{\rm eq}\sim1500\,{\rm K}$ and $T_{\rm int}\sim100\,{\rm K}$, irradiation pushes the boundary to hundreds of bars for typical increasing opacity. Jupiter has $T_{\rm eq}$ and $T_{\rm int}$ both of order $10^2\,{\rm K}$ and becomes convective near the bar scale. The estimate is order-of-magnitude because real opacities depend on both pressure and temperature.
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