$\Delta U = 0$

In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwiase.

$\therefore \,\,\,\Delta W =  - Area\,of\,rectangle\,ABCD =  - P\left( {2V} \right)$

$ =  - 2PV$

According to first law of thermodynamics

$\Delta Q = \Delta u + \Delta W\,or\,\Delta Q = \Delta W\,\,\left( {As\,\Delta u = 0} \right)$

$i.e.,$ heat supplied to the system is equal to the work done

So heat absorbed,$\Delta Q = \Delta W =  - 2PV$

$\therefore $ Heat rejected by the gas  $ = 2PV$