S _ 1 and S _ 2 are two identical sound sources of frequency 656Hz...

Given: ${f}_{0} = 656Hz$

$v = 324m{s}^{- 1}$

Frequency heard due to movement of $( {S}_{1} )$

${f}_{1} = \left(\dfrac{v}{v - {u}_{s}}\right){f}_{0}$

$⇒ {f}_{1} = \dfrac{324}{324 - 4} \times 656 = \dfrac{324}{320} \times 656$

And frequency heard due to movement of $( {S}_{2} )$ would be same as that of ${f}_{0}$, because the direction of velocity at position $R$ is perpendicular to the line $PR$, so there is no relative change in position at that moment.

${f}_{2} = 656Hz$

$∴$ Beat frequency $∆ f = {f}_{1} - {f}_{2} = 656\left(\dfrac{324}{320} - 1\right)$

$⇒ ∆ f = 8 . 2$