Step 1: Identify the path CA.
In this cycle the step CA happens at constant pressure $P_0$, with the volume changing from $2V_0$ at C down to $V_0$ at A. So CA is an isobaric (constant pressure) compression.
Step 2: Heat at constant pressure.
For an isobaric process the heat exchanged is $Q = nC_P\,\Delta T$, and for a diatomic gas $C_P = \frac{7R}{2}$.
Step 3: Find the temperature change using the gas law.
Since $PV = nRT$ and pressure is fixed at $P_0$, we get $nR\,\Delta T = P_0\,\Delta V = P_0(V_0 - 2V_0) = -P_0 V_0$.
Step 4: Substitute into the heat formula.
\[ Q_{CA} = n\cdot\frac{7R}{2}\,\Delta T = \frac{7}{2}\,(nR\,\Delta T) = \frac{7}{2}(-P_0V_0) \]
Step 5: Read the sign.
\[ Q_{CA} = -\frac{7}{2}P_0V_0 \]
The negative sign tells us heat leaves the gas.
Step 6: State the heat rejected.
The amount of heat given out to the surroundings is the magnitude, $\frac{7}{2}P_0V_0$. (The two-thirds mole value drops out because $nR\Delta T$ was found directly from $P_0\Delta V$.)
\[ \boxed{Q_{\text{rejected}} = \frac{7}{2}P_0V_0} \]