According to the question 
Speed of listener VL=1 ms−1 Speed of sound V=330 ms−1 Frequency of each source v=600 Hz ∴ Apparent frequency due to P=V′ =Vv( V−VL) Apparent frequency due to Q=V′′= Vv( V−VL) ∴ No. of heat heard by the listener per second. V′′−V′=Vv( V−VL)−Vv( V0−VL)=v2vL=3302×660×1=4