Step 1: Compare the two processes.
Both gases start the same and are squeezed to half the volume. Gas $A$ is squeezed slowly so its temperature stays fixed (isothermal). Gas $B$ is squeezed with no heat escaping (adiabatic), so it heats up.
Step 2: Pressure for the isothermal gas A.
For a fixed temperature, $PV$ stays constant. \[ P V = P_A' \left(\frac{V}{2}\right) \] So \[ P_A' = 2P \]
Step 3: Pressure for the adiabatic gas B.
Here $PV^\gamma$ stays constant, where $\gamma > 1$. \[ P V^\gamma = P_B' \left(\frac{V}{2}\right)^\gamma \] So \[ P_B' = P\,2^\gamma \]
Step 4: Compare the two factors.
Since $\gamma$ is bigger than $1$, the factor $2^\gamma$ is bigger than $2$.
Step 5: Decide which pressure is larger.
Because $2^\gamma > 2$, gas $B$ ends up at higher pressure than gas $A$. \[ P_B' > P_A' \]
Step 6: State the conclusion.
So the final pressure of $A$ is less than that of $B$. \[ \boxed{P_A' < P_B'} \]