The $P-V$ diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle is : 

$$\begin{gathered} A\to B ; dW = 0 ; \\ Q =∆U=\frac{3}{2}(2P_0 V_0 –P_0 V_0 ) =\frac{3}{2}{P}_0 V_0 \\ B\to C ; W = 2P_0 V_0 ; \\ ∆U = \frac{3}{2}(4P_0 V_0 – 2P_0 V_0 ) = 3P_0 V_0 ; Q = 5P_0 V_0 \\ C\to D ; W = 0; ∆U = –3P_0 V_0 ; Q = –3P_0 V_0 \\ D\to A ; W = –P_0 V_0 ; ∆U = –\frac{3}{2}{P}_0 V_0 ; Q = – \frac{5}{2}{P}_0 V_0 \\ so, W_{cycle} = P_0 V_0 and heat from source = {\left(Q\right)}_{+}= 6.5 P_0 V_0 \end{gathered}$$