The detector is fixed at point P, so only the motion of the source $$S_2$$ (real frequency $$f = 656\;\text{Hz}$$) produces the Doppler shift.

Geometry of the situation
Take the centre $$O$$ of the circle as the origin, the radius of the circle as $$r$$ and let the positive $$x$$-axis pass through P.
 • Coordinates: $$P(r,0),\;Q(0,r),\;R(-r,0)$$ (P and R are opposite, Q is equidistant).
 • At Q the position vector of $$S_2$$ is $$\vec{r}_Q=(0,r)$$.
 • Since $$S_2$$ moves anticlockwise with speed $$u = 4\sqrt2\;\text{m s}^{-1}$$, its velocity at Q is tangential and directed along negative $$x$$: $$\vec{u}_s = (-4\sqrt2,\,0)\;\text{m s}^{-1}$$.

Component of source velocity along the line Q → P
Vector from Q to P: $$\vec{QP} = (r,\, -r)$$.
Unit vector from Q to P: $$\hat{n} = \frac{\vec{QP}}{|\vec{QP}|}= \frac{(r,\,-r)}{r\sqrt2}= \left(\frac1{\sqrt2},\,-\frac1{\sqrt2}\right).$$
Radial component of the source velocity, $$u_r = \vec{u}_s \!\cdot\! \hat{n} = (-4\sqrt2,\,0)\!\cdot\!\left(\frac1{\sqrt2},\,-\frac1{\sqrt2}\right) = (-4\sqrt2)\!\left(\frac1{\sqrt2}\right)= -4\;\text{m s}^{-1}.$$ The negative sign means $$S_2$$ is moving \emph{away} from the detector along the line of sight at 4 m s$$^{-1}$$.

Doppler formula for a stationary observer and a moving source
If the source recedes with speed $$u_r$$, the observed frequency is $$f' = \frac{v}{\,v + |u_r|\,}\;f,$$ where $$v = 324\;\text{m s}^{-1}$$ is the speed of sound.

Substituting the values, $$f' = \frac{324}{324 + 4}\times 656 = \frac{324}{328}\times 656 = 0.987804878\times 656 = 648\;\text{Hz}.$$

Therefore, when only $$S_2$$ is emitting and it is at Q, the detector measures a frequency of 648 Hz.