For a cyclic process, the system returns to its initial state, so the change in internal energy ($\Delta U$) is zero . According to the First Law of Thermodynamics, $\Delta Q = \Delta U + \Delta W$, which implies $\Delta Q = \Delta W$ . The net work done ($\Delta W$) in a cyclic process is equal to the area enclosed by the cycle on the P-V diagram .

Assuming the standard figure for this AIPMT 2012 question (a rectangle with limits typically from $P$ to $2P$ and $V$ to $2V$, or similar dimensions yielding an area of $PV$): Area = $\Delta P \times \Delta V = (2P - P) \times (2V - V) = P \times V = PV$.

The question asks for heat rejected. In a cyclic process, if the cycle is traced counter-clockwise, work is done on the system (negative work), and heat is rejected by the system (negative heat). The magnitude of this heat is equal to the area of the loop, which is $PV$.