${K_P} = \frac{{{{({P_{N{H_3}}})}^2} \times ({P_{C{O_2}}})}}{{{P_{N{H_2}COON{H_4}(s)}}}}$

$ = {({P_{N{H_3}}})^2} \times ({P_{C{O_2}}})$

As evident by the reaction, $NH_3$ and $CO_2$ are formed in molar ratio of $2:1$. Thus if $P$ is the total pressure of the system at equilibrium, then

 ${P_{N{H_3}}} = \frac{{2 \times P}}{3}$ ${P_{C{O_2}}} = \frac{{1 \times P}}{3}$

${K_P} = {\left( {\frac{{2P}}{3}} \right)^2} \times \frac{P}{3} = \frac{{4{P^3}}}{{27}}$

Given ${K_P} = 2.9 \times {10^{ - 5}}$

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

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

$P = {\left( {\frac{{2.9 \times {{10}^{ - 5}} \times 27}}{4}} \right)^{1/3}} = 5.82 \times {10^{ - 2}}\,atm$