Solution:

a.
We show that not only is triangle $QAR$ isosceles with base $AR$ by construction, but triangle $PAQ$ is also isosceles (with base $AP$).

Indeed
$$B\widehat{A}P = P\widehat{A}Q$$
because $AP$ is a bisector, while
$$B\widehat{A}P = A\widehat{P}Q$$
because they are alternate interior angles with respect to the transversal $AP$ cutting the parallels $BA$ and $PQ$. Therefore also
$$P\widehat{A}Q = A\widehat{P}Q$$
that is, precisely, triangle $PAQ$ is isosceles. But the two isosceles triangles $QAR$ and $PAQ$ have the side $AQ$ in common, so
$$QP = QA = QR$$
which means that $Q$ is the circumcenter of triangle $APR$, as required.

b.
First we prove the following lemma.

Lemma 1. Triangle $APR$ is right-angled at $A$.

Proof.
Let us compute the measures of angles $P\widehat{A}Q$ and $Q\widehat{A}R$:
$$\begin{aligned} P\widehat{A}Q &= \frac{180^0 - P\widehat{Q}A}{2} \\ Q\widehat{A}R &= \frac{180^0 - A\widehat{Q}R}{2} \end{aligned}$$
but then
$$P\widehat{A}R = P\widehat{A}Q + Q\widehat{A}R = \frac{180^0 - P\widehat{Q}A}{2} + \frac{180^0 - A\widehat{Q}R}{2} = \frac{360^0 - (P\widehat{Q}A + A\widehat{Q}R)}{2} = 90^0$$
because $P\widehat{Q}A$ and $A\widehat{Q}R$ are supplementary (together they form the straight angle $P\widehat{Q}R$).

At this point it is easy to see that the points $P, A, R, T, S$ lie on the same circle centered at $Q$. Indeed, for $P, A$ and $R$ we already know this from the previous point; and moreover
- $S$ belongs to the same circle because $S$ and $A$ see the same segment $PR$ under a right angle (hence $APSR$ is a cyclic quadrilateral);

- $T$ also belongs to that circle, because its distance from the center $Q$, that is, the length of segment $QT$, equals the length of $QP$, which is a radius of the circle: indeed, since $PT$ is parallel to $AR$,
$$Q\widehat{P}T = Q\widehat{R}A$$
but on the other hand,
$$R\widehat{Q}A = P\widehat{Q}T$$
because they are vertical angles at $Q$, hence triangle $PQT$ is similar (in fact, congruent) to triangle $RQA$, which is isosceles with base $AR$, and is therefore also isosceles with base $PT$.

But then $APST$ is in turn a cyclic quadrilateral, so
$$P\widehat{A}T = 180^0 - P\widehat{S}T = T\widehat{S}C$$
from which we obtain that triangles $APC$ and $STC$ are indeed similar, having one angle in common and two corresponding equal angles (namely those at $A$ and at $S$ respectively).