1. Write the equilibrium reaction

$NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g)$

For this heterogeneous equilibrium, the solid does not appear in the equilibrium constant expression.

So,

$K_p = P_{NH_3}\cdot P_{H_2S}$

2. Initial condition

Initially, the flask already contains ammonia gas at pressure:

$P_{NH_3}^{\text{initial}} = 0.50\,\text{atm}$

Initially, no $H_2S$ is present:

$P_{H_2S}^{\text{initial}} = 0$

3. Let decomposition produce $x$ atm each of $NH_3$ and $H_2S$

From the reaction stoichiometry, decomposition of solid $NH_4HS$ adds equal partial pressures of $NH_3$ and $H_2S$.

Thus at equilibrium:

$P_{NH_3} = 0.50 + x$ $P_{H_2S} = x$

4. Use total equilibrium pressure

Given total pressure at equilibrium is:

$P_{\text{total}} = 0.84\,\text{atm}$

Therefore,

$(0.50 + x) + x = 0.84$

$0.50 + 2x = 0.84$

$2x = 0.34$

$x = 0.17\,\text{atm}$

So,

$P_{NH_3} = 0.50 + 0.17 = 0.67\,\text{atm}$ $P_{H_2S} = 0.17\,\text{atm}$

5. Calculate $K_p$

$K_p = P_{NH_3}\cdot P_{H_2S}$

$K_p = (0.67)(0.17) = 0.1139$

$K_p \approx 0.11$

6. Match with option

The correct option is:

$\boxed{0.11}$

So, Option B is correct.