= Solution
Take $S$ as the integrated line <mass absorption coefficient>, and $q$ as the absorber <mass fraction>, assumed constant; for a pure absorbing gas set $q=1$. All spectral intervals below use the paper's wavenumber coordinate. The normalized <Lorentz line shape> is
$$
\kappa_\nu(p)=\frac{S}{\pi}\frac{\Delta(p)}{(\nu-\nu_0)^2+\Delta(p)^2},\qquad
\boxed{\kappa_{\nu_0}(p)=\frac{S}{\pi\Delta(p)}.}
$$
If instead an absorption coefficient per length is wanted, multiply by absorber <density>. Neglecting temperature dependence, <pressure broadening> gives $\Delta(p)=\Delta_s p/p_s$. In the line wings, with $d=|\nu-\nu_0|\gg\Delta(p)$,
$$
\kappa_\nu\simeq\frac{S\Delta_s p}{\pi p_s d^2},\qquad
\tau_\nu(p)\simeq\frac{qS\Delta_s p^2}{2\pi gp_s d^2}.
$$
Define the line-opacity parameter and characteristic wavenumber width
$$
C=\frac{qSp_s}{2\pi g\Delta_s},\qquad w=\Delta_s\sqrt C.
$$
At an emitting level $\tau_\nu\sim1$, the wing estimate implies $d/\Delta(p)\sim\sqrt C$. Thus the <strong-line limit> relevant to outgoing radiation is
$$
\boxed{C\gg1,\quad\text{equivalently }qSp_s/(g\Delta_s)\gg1\text{ up to fixed factors}.}
$$
It ensures that the opaque-to-transparent transition lies in the Lorentz wings, rather than the narrow central core, and the characteristic absorption width is much larger than $\Delta_s$. At a grazing ray angle the criterion is even easier to satisfy because of the larger <optical depth>. This is a criterion for the radiatively important wings, not a claim that collision broadening remains physical at arbitrarily small <pressure>.
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