The idea is to show that the points $Z,B,E,P,T$ lie on the circle with diameter $ZT$.
I will start with a nice picture serving for illustration of the solution
and a not so nice comment. Composer of geometry problems often take a series of ten points introduced in a simple way, they get a concrete geometric constellation, then they find for the simplest points a (most) complicated way to introduce them. Each simple property becomes hard, except for the case when we also reverse the order. Terminology: "Composition by contorsion". Example: The point $P$ is simply the projection of $A$ on $CZ$. I will use $P'$ from the beginning for this projection, in a final it turns out that $P=P'$. (The "contorsion" is best combined with the choice of all letters from alphabet, so that the reader finds it hard to remember them and their properties.) This is a good way to produce "hard problems" e.g. for challenges, but is not a good way to structurally educate the young geometric eye and give it a direction of study.
I will break the solution into pieces. (Still keeping the notations for the parallel.)
Lemma:
In the right triangle $\Delta ACZ$, $\hat A=90^\circ$, let $CB$ be the median. Let $E$ be the projection of $A$ on $BC$. Set $R=AE\cap CZ$. Let $AP'$, $RL$, $CE$ be the heights in $\Delta ARC$, which intersect in its orthocenter $H$.
Then: $Z,E,L$ are colinear, $\widehat{ZER}=\widehat{ACZ}=\widehat{REP'}$, and $ER$ bisects $\widehat{ZEC}$.
Proof: From $RL\|ZA$ (right angles formed with $AC$), the reciprocal of the theorem of Ceva applied in $\Delta AZC$ starting from
$$
\underbrace{
\frac{LA}{LC}\cdot
\frac{RC}{RZ}}_{=1}\cdot
\underbrace{\frac{BZ}{BA}}_{=-1} = -1
$$
gives the concurrence of the cevians $CB$, $AR$, $ZL$. We consider the angles now and observe the relations:
$$
\widehat{ZER}
=
\widehat{EAZ}+
\widehat{EZA}
=
\widehat{ACB}+
\widehat{BCZ}
=
\widehat{ACZ}
\ .
$$
Indeed, $\widehat{EAZ}=\widehat{EAB}=90^\circ-\hat B= \widehat{ACB}$. The other equality of angles follows from the similarity $\Delta BZE\sim\Delta BCZ$. There is a common angle in $B$, and
$$
\frac{BZ}{BC}=
\frac{BE}{BZ}
$$
because of $BZ^2=AB^2=BC\cdot BE$. (The similitude "inside $\Delta BCZ$" is obtained by the similitude "inside $\Delta ACB$".)
$\square$
We come back to our problem.
Let $O$ be the center of $(c)$. The reflection in $O$ will be denoted by a star, so it is the map $X\to X^*$. For example, $F=D^*$.
Then $ACZA^*$ is a parallelogram. (Since $\widehat{CAZ}=90^\circ=\widehat{CAZ}$. Its diagonals intersect in $B$.) We know the two angles in $A^*$ in this parallelogram formed by the diagonal $ABCE$ with two sides, same as in $C$, so $AEZA^*$ cyclic, so $A^*$ is also on the circle $(c)$, its center $O$ is the mid point of $AA^*$ (because of the right angle in $E$).
Also $F=D^*$ is the point making $ADA^*F$ a parallelogram.
Let $Z^*$ be ($Z$ reflected in $O$). Then $ZAZ^*A^*$ is also a rectangle, and $AZ^*\|BO\|ZA^*$, and obtain $AZ^*=ZA^*=CA$.
Let us show that $F,E,P'$ are colinear. We show $\widehat{ACZ}=\hat C=\widehat{REP'}$. (The colinearity follows since $AER$ is a line.) We compute:
$$
\widehat{AEF} =
\widehat{AZF} =
90^\circ-\widehat{AZC} =
\widehat{ACZ} = \hat C
\widehat{REP'}
\ .
$$
Let $\gamma$ be the circle $\gamma = \odot(EBZ)$. Its center is denoted by $O'$.
Because of
$$
\frac 12\overset\frown{BZ}=
\widehat{BEZ} =
\widehat{A^*EZ} =
\widehat{A^*AZ} =
\widehat{Z^*ZA} =
\widehat{OZA}
$$
the line $OZ$ is tangent in $Z$ to $\gamma$. So the two circles $c,\gamma$ intersect orthogonally in $Z$ (and $E$). The point $P=P'$ is also on $\gamma$ because of
$$
\widehat{EPZ} =
\widehat{EPR} =
\widehat{RAC} =
\widehat{EAC} =
\widehat{CBA} =
\widehat{EBA}
\ ,
$$
so $PEBZ$ cyclic.
Finally, since $\widehat{ZPA}$ is a right angle, the point $T$ making $ZT$ a diameter of $\gamma$ is on $PA=PA'$.
$\square$
Best Answer
Note that $\angle APB = \angle CNA$ and $\angle BAP = \angle NAC$. This shows that $\triangle APB \sim \triangle ANC$. Similarly, $\angle AMB = \angle CQA$ and $\angle BAM = \angle QAC$, hence $\triangle AMB \sim \triangle AQC$.
We have $$\angle PMN = \angle PMA = \angle PBA = \angle ACN = \angle AQN = \angle PQN,$$ hence $P,Q,M,N$ are concyclic.
Let $BM$ intersect the circumcircle of $PQMN$ again at $T\neq M$. Note that $\angle PTM = \angle PQM$ and $\angle MAP = \angle MBP$, hence $\triangle BPT \sim \triangle AMQ$. Therefore $\dfrac{BT}{AQ}=\dfrac{BP}{AM}$. On the other hand, similarity $\triangle AMB \sim \triangle AQC$ gives $\dfrac{BM}{AM}=\dfrac{CQ}{AQ}$. Hence the power of $B$ with respect to the circumcircle of $PQMN$ is equal to $$BM \cdot BT = \frac{AM\cdot CQ}{AQ} \cdot \frac{BP \cdot AQ}{AM} = CQ \cdot BP.$$
Similarly, let $CQ$ intersect the circumcircle of $PQMN$ again at $U\neq Q$. We have $\angle APN = \angle CUN$ and $\angle NAP = \angle NCU$, hence $\triangle CUN \sim \triangle APN$ which gives $\dfrac{CU}{CN}=\dfrac{AP}{AN}$. Moreover, $\triangle APB \sim \triangle ANC$ gives $\dfrac{CN}{AN}=\dfrac{BP}{AP}$. Therefore the power of $C$ with respect to the circumcircle of $PQMN$ is equal to $$CQ\cdot CU = CQ \cdot \frac{CN\cdot AP}{AN} = CQ \cdot \frac{\frac{BP\cdot AN}{AP}\cdot AP}{AN} = CQ \cdot BP.$$
This shows that $B$ and $C$ have equal powers with respect to the circumcircle of $PQMN$. Denote the center and the radius of this circle by $O$ and $r$. Then $BO^2-r^2=CO^2-r^2$ from which it follows that $BO=CO$. In other words, $O$ lies on the perpendicular bisector of $BC$.