Step-by-step Solution:

1. Understand the problem setup and identify given values:

   • A stationary sound detector is at point $P$.
   • A sound source $S_2$ moves anti-clockwise on a circular path around center $O$.
   • The detector $P$ is also on this path.
   • Points $P$ and $R$ are diametrically opposite, and point $Q$ is equidistant from $P$ and $R$. This geometric arrangement implies that the angle $\angle POQ$ is $90^\circ$, and $\triangle POQ$ is a right-angled isosceles triangle.
   • Frequency of the source, $f_0 = 656 \mathrm{~Hz}$.
   • Speed of the source, $v_s = 4\sqrt{2} \mathrm{~m/s}$.
   • Speed of sound in air, $v = 324 \mathrm{~m/s}$.
   • We need to find the frequency measured by the detector at $P$ when the source $S_2$ is at $Q$.

2. Apply the Doppler Effect formula: The general formula for the apparent frequency $f'$ heard by a stationary observer from a moving source is: $f' = f_0 \left( \frac{v}{v - v_{||}} \right)$ where $v_{||}$ is the component of the source's velocity along the line connecting the source to the observer. This component is taken as positive if the source is moving towards the observer and negative if it is moving away.

3. Determine the geometry and the velocity component $v_{||}$:

   • Let the center of the circle $O$ be the origin $(0,0)$. Let the radius of the circle be $r$.
   • Let the position of the detector $P$ be $(r, 0)$.
   • Since $Q$ is on the circle and $\angle POQ = 90^\circ$ (due to anti-clockwise motion from $P$), the position of the source $S_2$ at $Q$ is $(0, r)$.
   • The velocity of the source $S_2$ is always tangent to the circular path. At point $Q(0, r)$, for anti-clockwise motion, the velocity vector $\vec{v}_s$ is directed along the negative x-axis. So, $\vec{v}_s = (-v_s, 0)$.
   • The line connecting the source $S_2$ (at $Q$) to the observer (at $P$) is represented by the vector $\vec{QP} = \vec{P} - \vec{Q} = (r, 0) - (0, r) = (r, -r)$.
   • The component of the source's velocity along the line of sight ($QP$) is the projection of $\vec{v}_s$ onto $\vec{QP}$. Let $\theta$ be the angle between $\vec{v}_s$ and $\vec{QP}$. Then $v_{||} = v_s \cos\theta$.
   • Alternatively, we can use the dot product: $v_{||} = \vec{v}_s \cdot \hat{u}_{QP}$, where $\hat{u}_{QP}$ is the unit vector from $Q$ to $P$. $\hat{u}_{QP} = \frac{\vec{QP}}{|\vec{QP}|} = \frac{(r, -r)}{\sqrt{r^2 + (-r)^2}} = \frac{(r, -r)}{r\sqrt{2}} = \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$
   • Now, we calculate $v_{||}$: $v_{||} = (-v_s, 0) \cdot \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) = -v_s \left(\frac{1}{\sqrt{2}}\right) + 0 = -\frac{v_s}{\sqrt{2}}$
   • The negative sign indicates that the source is effectively moving away from the observer along the line of sight.

4. Calculate the numerical value of $v_{||}$:

   • Given $v_s = 4\sqrt{2} \mathrm{~m/s}$.
   • $v_{||} = -\frac{4\sqrt{2}}{\sqrt{2}} = -4 \mathrm{~m/s}$.

5. Calculate the apparent frequency $f'$:

   • Substitute the values into the Doppler formula: $f' = f_0 \left( \frac{v}{v - v_{||}} \right) = 656 \left( \frac{324}{324 - (-4)} \right)$ $f' = 656 \left( \frac{324}{324 + 4} \right) = 656 \left( \frac{324}{328} \right)$
   • Simplify the fraction: $\frac{324}{328} = \frac{324 \div 4}{328 \div 4} = \frac{81}{82}$
   • Now, calculate $f'$: $f' = 656 \times \frac{81}{82}$
   • We notice that $656 = 8 \times 82$. $f' = (8 \times 82) \times \frac{81}{82} = 8 \times 81$ $f' = 648 \mathrm{~Hz}$

Final Answer: The frequency of sound measured by the detector is $648 \mathrm{~Hz}$.