Step-by-step Solution:

1. Analyze the setup and identify given values:

   • Frequency of both sources, $f_0 = 656 \mathrm{~Hz}$.
   • Speed of sound in air, $v = 324 \mathrm{~m} \mathrm{~s}^{-1}$.
   • The detector is stationary at point $P$.
   • Source $S_1$ is moving from the center $O$ towards the detector at $P$ with speed $v_{S_1} = 4 \mathrm{~m} \mathrm{~s}^{-1}$.
   • Source $S_2$ is at point $R$, which is diametrically opposite to $P$.

2. Calculate the apparent frequency from source $S_1$ ($f_1$):

   • Source $S_1$ is moving towards the stationary detector.
   • We use the Doppler effect formula for a source moving towards a stationary observer: $f' = f_0 \left( \frac{v}{v - v_s} \right)$.
   • Here, $f_0 = 656 \mathrm{~Hz}$, $v = 324 \mathrm{~m} \mathrm{~s}^{-1}$, and $v_s = v_{S_1} = 4 \mathrm{~m} \mathrm{~s}^{-1}$.
   • Substituting the values: $f_1 = 656 \left( \frac{324}{324 - 4} \right) = 656 \left( \frac{324}{320} \right) \mathrm{~Hz}$

3. Calculate the apparent frequency from source $S_2$ ($f_2$):

   • Source $S_2$ is at point $R$ and moving on a circular path around the center $O$. The detector is at point $P$.
   • Since $P$ and $R$ are diametrically opposite, the line connecting the source $S_2$ (at $R$) and the detector (at $P$) is the diameter $RP$.
   • The velocity of $S_2$ is always tangential to the circular path. At point $R$, the velocity vector $\vec{v}_{S_2}$ is perpendicular to the radius $OR$.
   • The line of sight from the source $S_2$ to the detector at $P$ is the line $RP$, which contains the radius $OR$.
   • Therefore, the velocity vector of $S_2$ is perpendicular to the line of sight between the source and the detector.
   • The component of the source's velocity along the line of sight is zero ($v_{S_2, \text{radial}} = v_{S_2} \cos(90^\circ) = 0$).
   • Since there is no relative velocity component along the line joining the source and observer, there is no change in frequency due to the Doppler effect.
   • So, the apparent frequency detected from $S_2$ is equal to its original frequency: $f_2 = f_0 = 656 \mathrm{~Hz}$

4. Calculate the beat frequency ($f_{\text{beat}}$):

   • The beat frequency is the absolute difference between the two frequencies detected. $f_{\text{beat}} = |f_1 - f_2|$
   • Substituting the expressions for $f_1$ and $f_2$: $f_{\text{beat}} = \left| 656 \left( \frac{324}{320} \right) - 656 \right|$
   • Factor out 656: $f_{\text{beat}} = 656 \left| \frac{324}{320} - 1 \right|$
   • Simplify the expression inside the absolute value: $f_{\text{beat}} = 656 \left| \frac{324 - 320}{320} \right| = 656 \left( \frac{4}{320} \right)$
   • Perform the final calculation: $f_{\text{beat}} = 656 \left( \frac{1}{80} \right) = \frac{656}{80} = \frac{65.6}{8} = 8.2 \mathrm{~Hz}$

The beat frequency measured by the detector is $8.2 \mathrm{~Hz}$.