A monoatomic gas at pressure P and volume V is suddenly compressed...

Change in entropy is given by $d S = \dfrac{d Q}{T}$. Since, the process is $d S = 0$, $d Q = 0$. Therefore, constant entropy means the process is adiabatic.

$P{V}^{\gamma} =$constant

$⇒ {P}_{1}{V}_{1}^{\gamma} = {P}_{2}{V}_{2}^{\gamma}$

For a monatomic gas $\gamma = \dfrac{5}{3}$

$⇒ {P}_{1}{V}_{1}^{\dfrac{5}{3}} = {P}_{2}\left(\dfrac{{V}_{1}}{8}\right)\dfrac{5}{3}$

$⇒ {P}_{1}{V}_{1}^{\dfrac{5}{3}} = \dfrac{{P}_{2}{V}_{1}^{\dfrac{5}{3}}}{32}$

$⇒ {P}_{2} = 32{P}_{1}$