Consider the heterogeneous equilibrium in a closed container

${\mathrm{NH}}_4\mathrm{HS}\left(\mathrm{s}\right) \rightleftharpoons {\mathrm{NH}}_3\left(\mathrm{g}\right) + {\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)$

If more ${\mathrm{NH}}_4\mathrm{HS}$ is added to the equilibrium

${\mathrm{NH}}_4\mathrm{HS}\left(\mathrm{s}\right) \rightleftharpoons {\mathrm{NH}}_3\left(\mathrm{g}\right) + {\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)$

${\mathrm{K}}_{\mathrm{P}}={\mathrm{NH}}_3\left(\mathrm{g}\right) \times {\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)$

Partial pressures of ${\mathrm{NH}}_3$ and ${\mathrm{H}}_2\mathrm{O}$ do not affect ${\mathrm{NH}}_4\mathrm{HS}$(s) because ${\mathrm{NH}}_4\mathrm{HS}$(s) is solid for solid active mass unity.