$2N_2{O}_5\left(g\right) \to 4N{O}_2\left(g\right)+O_2\left(g\right)$

at t=0,            P₀                     0              0

at t=30 min.   (P₀-x)                2x            $\frac{x}{2}$

at t=60 min.   (P₀-y)                2y            $\frac{y}{2}$

$\because P_0=50 mmHg$

$\because after 30min.$  $⟹\mathrm{ }\left({\mathrm{P}}_0-\mathrm{x}\right)+2\mathrm{x}+\frac{\mathrm{x}}{2}=87.5$

$P_0+\frac{3}{2}x=87.5$

$\frac{3}{2}x=37.5 ; x=25mmHg$

$P_0=50 ; P_{30min}=25mmHg$

That mean 30 min. is Half life of reaction. 

$P_0\stackrel{30\mathrm{m}\mathrm{i}\mathrm{n}.}{\to }\frac{P_0}{2}\stackrel{30\mathrm{m}\mathrm{i}\mathrm{n}.}{\to }\frac{P_0}{4}$

at t= 60 min

$P_{N_2{O}_{5 }}=\frac{P_0}{4} ⟹\frac{50}{4}=12.5mmHg$

$P_0-y=12.5 ⟹50-y=12.5$

$y=37.5mmHg$

Total P after 60 min. $⟹\left(P_0-y\right)+2y+\frac{y}{2}⟹50\times \frac{3}{2} \times 37.5$

$=106.25 mmHg$