As per law of mass action, for ${\mathrm{NH}}_4{\mathrm{HS}}_{\left(\mathrm{s}\right)}\rightleftharpoons$${\mathrm{NH}}_{3( \mathrm{g})}+{\mathrm{H}}_2{ \mathrm{S}}_{\left(\mathrm{g}\right)}$ ratio of rate constants of forward $\left({\mathrm{K}}_{\mathrm{f}}\right)\&$ backward $\left({\mathrm{K}}_{\mathrm{b}}\right)$ reactions at equilibrium equals to

As per law of Mass Action?

${\mathrm{NH}}_{4}HS\left(S\right)\rightleftharpoons {\mathrm{NH}}_3\left(g\right)+{\mathrm{H}}_2 \mathrm{S}\left(g\right)$

$\therefore K_{eq}$  =$\frac{k_P}{K_b}=\frac{\text{ rate of forward reaction }}{\text{ rate of back ward reaction }}$

The solid state compounds are not considered for concentration,

$\therefore [$ [N HHs]$=1$.

$\mathrm{Keq}=\frac{\left[{\mathrm{NH}}_3\right]\left[{\mathrm{H}}_2 \mathrm{S}\right]}{1}=\left[{\mathrm{NH}}_3\right]\left[{\mathrm{H}}_2 \mathrm{S}\right]$

Hence the correct answer is (D) $\left[{\mathrm{NH}}_3\right]\left[{\mathrm{H}}_2 \mathrm{S}\right]$