A monoatomic gas is compressed to (1/8) th of its volume...

For an adiabatic process, the relation between pressure and volume is given by $PV^{\gamma} = \text{constant}$.

For a monoatomic gas, the ratio of specific heats is $\gamma = \dfrac{5}{3}$.

Given that the initial volume is $V_1 = V$ and the final volume is $V_2 = \dfrac{V}{8}$.

Using the adiabatic equation, $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$

$P_0 V^{\dfrac{5}{3}} = P_2 \left(\dfrac{V}{8}\right)^{\dfrac{5}{3}}$

$P_2 = P_0 (8)^{\dfrac{5}{3}} = P_0 (2^3)^{\dfrac{5}{3}} = P_0 (2^5) = 32 P_0$

The increase in pressure is $\Delta P = P_2 - P_1$

$\Delta P = 32 P_0 - P_0 = 31 P_0$

Answer: $31\ P_0$