a. We know that triangle $ABC$ is an equilateral. Thus we have $\widehat{PRQ} = \widehat{PQR} = 90^\circ - 60^\circ = 30^\circ$ and by this we have $PQ = PR$. Suppose that $S$ is the circumcenter of triangle $PQR$.

We have $\widehat{QPR} = 180^\circ - 2 \times 30^\circ = 120^\circ$. So $\widehat{QR} = 120^\circ$ and moreover $\widehat{QSR} = 120^\circ$. $SQ = SR$, so $\widehat{SQR} = \widehat{SRQ}$. By these arguments we conclude that triangle $SQR$ is the reflection of triangle $PQR$ respect to $BC$. So $S$ is the reflection of $P$ respect to $BC$. Thus $S$ moves on a circle which is the reflection of the circumcircle of triangle $ABC$ respect to $BC$. $\square$

b. Suppose that $S$ is the circumcenter of triangle $PQR$ and $O$ is the circumcenter of triangle $ABC$. By considering $P$ on $B, C$ we can obviously see that $S$ lies on $B, C$ respectively. Now we move $P$ to the intersection point of $AO$ and the circumcircle of triangle $ABC$. In this case we have $\widehat{PCA} = \widehat{PBA} = 90^\circ$. So $S$ lies on $O$. The parallel line from $P$ respect to $BC$ meets the circumcircle of triangle $ABC$ at $P'$. By considering $P'$ as $P$, triangle $PQR$ is similar to the triangle $PQR$ in the previous case. Note that the $P$ altitudes in both of these triangles have equal lengths. So these triangles are equivalent. Thus by transferring $O$ by $pp'$ which is parallel to $BC$, we reach the point $S$, the circumcenter of triangle $PQR$ in this case. We know that $BSOC$ is cyclic. But this can't be true unless $S$ lies on $O$. It yields that $pp'$ must be equal to 0. So $P'$ lies on $P$ and triangle $ABC$ is isosceles with $AB = AC$. Now let $P$ lie on $A$.

$\omega$ is the circumcircle of triangle $BOC$. Triangle $ABC$ is isosceles. So the perpendicular bisector of $RQ$ is the $A$ altitude in triangle $ABC$. This perpendicular bisector meets $\omega$ for the second time at $S$. We know that the circumcenter of triangle $AQR$ lies on $\omega$, so it's whether $O$ or $S$. It can't be $O$. Because otherwise we have
$$OB = OA = OR$$
Which can't be true. So $S$ is the circumcenter of triangle $AQR$. We have
$$\left. \begin{array}{l} SA = SQ \implies \widehat{SAQ} = \widehat{SQA} \\[1.5ex] \widehat{CAQ} = \widehat{BAQ} - \widehat{BAC} = 90^\circ - \hat{A} \\[1.5ex] \widehat{SAC} = \frac{\hat{A}}{2} \end{array} \right\} \implies \widehat{SAQ} = 90^\circ - \frac{\hat{A}}{2}$$
By these we have
$$\begin{align*} \widehat{SQA} &= \widehat{SAQ} = 90^\circ - \frac{\widehat{A}}{2} &\implies \widehat{ASQ} &= 180^\circ - 2 \times (90^\circ - \frac{\widehat{A}}{2}) = \widehat{A} \\ \implies \widehat{CQA} &= 90^\circ - \widehat{ASQ} = 90^\circ - \widehat{A}, \widehat{AQS} = 90^\circ - \frac{\widehat{A}}{2} &\implies \widehat{AQC} &= \frac{\widehat{A}}{2} \end{align*}$$
Moreover we have
$$\widehat{ACD} = 180^\circ - 90^\circ - (90^\circ - \widehat{B}) = \widehat{B}$$
In the circumcircle of triangle $ACQ$ we have
$$\widehat{CAS} = \frac{\widehat{A}}{2} = \widehat{AQC} = \frac{\widehat{AC}}{2}$$

$$\widehat{CQS} = 90^\circ - \widehat{A} = \widehat{CAQ} = \frac{\widehat{CQ}}{2}$$
Using these arguments we conclude that $SA$ and $SQ$ are tangent to the circumcircle of triangle $ACQ$. Let $O_1$ be the circumcenter of triangle $ACQ$. Then we have

$$\widehat{CO_1A} = \widehat{AC} = 2\widehat{CQA} = \widehat{A} \\ \widehat{CAO_1} = \widehat{ACO_1} = \frac{180^\circ - \widehat{CO_1A}}{2} = 90^\circ - \frac{\widehat{A}}{2}$$
We knew that $\widehat{ACD} = \hat{B} = 90^\circ - \frac{\hat{A}}{2}$. So $\widehat{ACD} = \widehat{ACO_1}$. Thus it yields that $C$ lies on $AO_1$ and $AC = CQ$. So we have
$$\begin{align*} \widehat{CAQ} = \widehat{CQA} &\implies 90^\circ - \widehat{A} = \frac{\widehat{A}}{2} &\implies \frac{3}{2}\widehat{A} &= 90^\circ \\ &\implies \widehat{A} = 60^\circ \quad \square \end{align*}$$