A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is

Area of the upper triangle
$=\frac{1}{2}\times \left(2{\mathrm{V}}_0-{\mathrm{V}}_0\right)\times \left(3{\mathrm{P}}_0-2{\mathrm{P}}_0\right)=\frac{{\mathrm{P}}_0{\mathrm{V}}_0}{2}$
So work done in that cycle
${\mathrm{W}}_1=\frac{-{\mathrm{P}}_0{\mathrm{V}}_0}{2}$[Anticlockwise]
Area of the lower triangle
$=\frac{1}{2}\times \left(2{\mathrm{V}}_0-{\mathrm{V}}_0\right)\times \left(2{\mathrm{P}}_0-{\mathrm{P}}_0\right)=\frac{{\mathrm{P}}_0{\mathrm{V}}_0}{2}$
So work done in this cycle
${\mathrm{W}}_2=\frac{+{\mathrm{P}}_0{\mathrm{V}}_0}{2} [\mathrm{Clockwise}]$
Total workdone$\mathrm{W}={\mathrm{W}}_1+{\mathrm{W}}_2=0$