b
\(N{H_2}COON{H_4}(s) \leftrightarrow 2N{H_3}(g) + C{O_2}(g)\)

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

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

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 \({K_P} = 2.9 \times {10^{ - 5}}\)

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

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

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