Equilibrium $\equiv$ Rate Of Forward $=$ Backward Reaction

Since $\ce{NH4COONH2}$ is a solid, it does not contribute to the pressure in the vessel. The total pressure of the vessel can be written as the sum of the partial pressures of $\ce{NH3}$ and $\ce{CO2}$ .

Initially, since there is no $\ce{NH3}$ or $\ce{CO2}$ in the vessel, the total pressure is zero. Then, $\ce{NH4COONH2}$ dissociates, and $\ce{NH3}$ and $\ce{CO2}$ are formed in the ratio $2: 1$ . Thus, their partial pressures are also in the ratio $2: 1$ . If the partial pressure of $\ce{CO2}$ is $P_{i}$ , then the total pressure in the vessel, $P_1$ , is $2P_{i} + P_{i} = 3P_{i}$

Then, after $\ce{NH3}$ is added to the vessel and equilibrium is attained again, the new partial pressure of $\ce{NH3}$ is $3 P_i$ . Let the new partial pressure of $\ce{CO2}$ be $P_{f}$ . The new total pressure $P_2$ , is $3P_i + P_f$ .

This data is summarized in the following table:

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\large \ce{NH4COONH2_{~~(s)} \rightleftharpoons 2NH3_{~~(g)} + CO2_{~~(g)}}$

$\underline{\text{Initial:}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 0 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 0~~~~~~~~~~~~~~~~~~0 \\ \underline{\text{First equilibrium:}}~~~~~~~~~~~~~~~~~~~~~~~~ 0 ~~~~~~~~~~~~~~~~~~~~~~~~~~ 2P_{i} ~~~~~~~~~~~~~~ P_{i}~~~~~~~~~~~~~~~~~~~~\small \color{#302B94}{P_{i} ~~\text{represents initial partial pressure of} ~~\ce{CO2}} \\ \underline{\text{Second equilibrium:}}~~~~~~~~~~~~~~~~~~~~ 0 ~~~~~~~~~~~~~~~~~~~~~~~~~~ 3P_{i} ~~~~~~~~~~~~~~~ P_{f}~~~~~~~~~~~~~~~~~~~\small \color{#302B94}{P_{f} ~~\text{represents final partial pressure of} ~~\ce{CO2}}$

Since the temperature does not change, there is no change in the equilibrium constant for this reaction.

$\therefore K_{P} = {(2P_{i})}^{2}(P_{i}) = {(3P_{i})}^{2}(P_{f}) \Longrightarrow P_{f} = \dfrac{4}{9}P_{i} ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \small \color{#302B94}{\text{where}~K_{P}~\text{represents value of equilibrium constant in terms of pressure}}$

Hence,

$\text{Required ratio} = \dfrac{3P_{i}+P_{f}}{2P_{i}+P_{i}} = \dfrac{3P_{i}+\dfrac{4}{9}P_{i}}{3P_{i}} = \boxed{\dfrac{31}{27}}$