For a reaction, \({N}_{2}{O}_{5(g)}\to 2N{O}_{2(g)}+\frac{1}{2}{O}_{2(g)}\) in a constant volume container, no products …

\({N}_{2}{O}_{5}(g)\to 2N{O}_{2}(g)+\frac{1}{2}{O}_{2}(g)\)

Let the initial pressure of \({N}_{2}{O}_{5}\) be \({P}_{0}\).
Initial moles: Only \({N}_{2}{O}_{5}\) is present \(\to {n}_{0}\) \(={P}_{0}\) (as per ideal gas law).
Let x be the fraction of reaction completed.
For 1 mole of \({N}_{2}{O}_{5}\) decomposing:
2 moles of \(N{O}_{2}\) are formed.
0.5 moles of \({O}_{2}\) are formed.

Given that 50 % of the reaction is completed (x =0.5):
\({N}_{2}{O}_{5}\) decomposed: \(0.5{P}_{0}\)
\(N{O}_{2}\) formed: \(2\times 0.5{P}_{0}={P}_{0}\)
\({O}_{2}\) formed: \(0.5\times 0.5{P}_{0}=0.25{P}_{0}\)
Remaining \({N}_{2}{O}_{5}:{P}_{0}-0.5{P}_{0}=0.5{P}_{0}\)
\({P}_{\text{final }}={P}_{{N}_{2}{O}_{5}}+{P}_{N{O}_{2}}+{P}_{{O}_{2}}\)
\({P}_{\text{final }}=0.5{P}_{0}+{P}_{0}+0.25{P}_{0}\)\(=1.75{P}_{0}=\frac{7}{4}{P}_{0}\)