For the decomposition reaction

${\mathrm{NH}}_2{\mathrm{COONH}}_4\left(\mathrm{s}\right) \leftrightharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right) + {\mathrm{CO}}_2\left(\mathrm{g}\right)$

The ${\mathrm{K}}_{\mathrm{P}}=2.9 \times {10}^{-5} {\mathrm{atm}}^3.$ The total pressure of gases at equilibrium when 1 mole of ${\mathrm{NH}}_2{\mathrm{COONH}}_4\left(\mathrm{s}\right)$ was taken to start with would be

${\mathrm{NH}}_2{\mathrm{COONH}}_4\left(\mathrm{s}\right) \leftrightharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right) + {\mathrm{CO}}_2\left(\mathrm{g}\right)$

         1                             2                 1

${\mathrm{K}}_{\mathrm{P}}=2.9 \times {10}^{-5} {\mathrm{atm}}^3$

If the P is the total pressure at equilibrium

${\mathrm{K}}_{\mathrm{P}}={\left(\frac{2\mathrm{p}}{3}\right)}^2 \left(\frac{\mathrm{p}}{3}\right)$

$\therefore {\mathrm{P}}^3=\frac{27\times 2.9\times {10}^{-5}}{4}=1.9575$

$\mathrm{P}=\sqrt[3]{1.9575}=0.058$