We are given a circle with center $O$. $SOQ$ is a diameter of the circle. We are given that $\angle SRP = 37^\circ$. We need to find $\angle PSQ$.

1. Problem Description

We are given a circle with center $O$. $SOQ$ is a diameter of the circle. We are given that $\angle SRP = 37^\circ$. We need to find $\angle PSQ$.

2. Solution Steps

Since $SOQ$ is a diameter, $\angle SQP$ is an angle subtended by a diameter. Therefore, $\angle SQP = 90^\circ$.

Angles subtended by the same arc are equal.

Since $\angle SRP$ and $\angle SQP$ are angles subtended by the chord $SP$,

$\angle SRP = \angle SQP$.

However, $\angle SRP$ is given to be $37^\circ$. So $\angle SQP$ should also be $37^\circ$. The argument above contains an error and the relationship between the angles are incorrect.

The correct interpretation is $\angle SRP$ and $\angle SQR$ are angles subtended by the same chord SR. Thus $\angle SQR = \angle SRP = 37^{\circ}$.

Since $SOQ$ is a diameter, $\angle QSP = 90^{\circ}$. Thus, triangle $QSP$ is a right triangle. The angles in a triangle add up to $180^\circ$.

Now, consider triangle $SQP$.

$\angle SQP + \angle QPS + \angle PSQ = 180^{\circ}$.

We also know that $\angle SQP = 90^\circ$. Therefore $\angle QSP + \angle QPS = 90^\circ$.

We also have the fact that $\angle SQR = 37^\circ$.

In the triangle $SQP$, since $\angle SQP$ is a right angle, we have that $\angle PSQ + \angle SPQ = 90^{\circ}$.

$\angle PSQ = 90^{\circ} - \angle SPQ$.

We need to figure out $\angle SPQ$.

Notice that $\angle SPQ = \angle SQP + \angle PQS$.

Since $\angle QSP = 90^{\circ}$, we have $\angle SQR = 37^{\circ}$. We also have $\angle SQP = 90^\circ$, and $QSP$ is a straight line.

The angle $\angle PSQ$ we want to find is part of triangle $QSP$ which is a right triangle where $\angle SQP = 90^\circ$. Also $\angle SQR = 37^\circ$ and $\angle QSP = 90^{\circ}$.

Therefore $\angle PSQ = \angle PSQ$.

Let's consider the angles $\angle RSQ$ and $\angle RPQ$ subtended by the chord $RQ$. These two angles are equal.

Also, consider $\angle RQS$ and $\angle RPS$. These two are equal.

Since $SOQ$ is a diameter, $\angle RSP = 90^\circ$.

Then $\angle RSP = \angle RSA + \angle PSA = 90^\circ$.

Also, since $\angle SRP = 37^\circ$, we have $\angle PSQ = 90^{\circ} - 37^\circ$. Then $\angle PSR + \angle PSQ = 90^\circ + \angle PSQ$

$\angle RPS$ = angles subtended by chord $QS = \angle RQS$ = $37^\circ$

$\angle PSQ = 53^\circ$

3. Final Answer

The final answer is C.