For ${\mathrm{NH}}_4\mathrm{HS}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{NH}}_3\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)$ the observed pressure for reaction mixture in equilibrium is 1.12 atm at 106°c. The value of K, for the reaction is:

$\underset{ \mathrm{Pressure} \mathrm{at} \mathrm{equilibrium} }{{\mathrm{NH}}_4\mathrm{HS}\left(\mathrm{s}\right)}\rightleftharpoons \underset{\mathrm{P}}{{\mathrm{NH}}_3\left(\mathrm{g}\right)}+\underset{\mathrm{P}}{{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)}$
$\therefore$ Total pressure at equilibrium = 2p = 1. 12 atm
P = 1.12/2 atm
$\therefore$ ${\mathrm{K}}_{\mathrm{p}}={\mathrm{P}}_{{\mathrm{NH}}_3}^{\mathrm{\prime }}\times {\mathrm{P}}_{{\mathrm{H}}_2\mathrm{S}}^{\mathrm{\prime }}$
$\therefore {\mathrm{K}}_{\mathrm{p}}=\frac{1.12}{2}\times \frac{1.12}{2}=0.3136{\mathrm{atm}}^2$