1. Identify the process

The phrase "at constant entropy" means the compression is adiabatic and reversible, i.e. an isentropic process.

For an ideal gas in an isentropic process,

$PV^\gamma = \text{constant}$

For a monoatomic gas,

$\gamma = \frac{C_P}{C_V} = \frac{5}{3}$

2. Use the given volume change

Initial state:

$P_1 = P, \quad V_1 = V$

Final volume is one eighth of original:

$V_2 = \frac{V}{8}$

Using

$P_1 V_1^\gamma = P_2 V_2^\gamma$

we get

$P \cdot V^{5/3} = P_2 \left(\frac{V}{8}\right)^{5/3}$

3. Solve for final pressure

$P_2 = P \cdot \frac{V^{5/3}}{(V/8)^{5/3}}$

$P_2 = P \cdot \left(\frac{V}{V/8}\right)^{5/3}$

$P_2 = P \cdot 8^{5/3}$

Now,

$8^{5/3} = (2^3)^{5/3} = 2^5 = 32$

So,

$P_2 = 32P$

4. Check options

• A: $P$ ❌
• B: $8P$ ❌
• C: $32P$ ✅
• D: $64P$ ❌

5. Final answer

The final pressure is

$\boxed{32P}$

So the correct option is C.