The fact that $A'$ must be the reflection of $A$ wrt $BC$ is useless, solution works for any point on $A$-altitude. Let $F$ be the foot of $A$-altitude, we claim that $P$ lies on $(FQS)$. Let $H$ be the point on $(FQS)$ so that $FHQS$ is isosceles trapezoid. Let $K$ be the reflection of $B$ over $B$, let $B'$, $C'$ be the reflections of $B$, $C$ over $F$, respectively. By the homothety, $B'KEC$ is cyclic. Let $I$, $L$ be the reflections of $K,E$ over line $AF$. Then $BILC'$ is cyclic. Also note that $BCEI$ and $BCKL$ are parallelograms, since $BL\parallel CK$ and $BI\parallel CE$, and $L-I-K-E\parallel BC$. Now, $H$ is the midpoint of $CL$, $Q$ the midpoint of $CI$. Also, by radical axis on $(B'KEC)$, $(BECD)$ and $(BLIC')$ implies that $D$ lies on $(BLIC')$, all this is enough to prove that by the homothety $P$ lies on $(FHQS)$.

rafaello wrote: The fact that $A'$ must be the reflection of $A$ wrt $BC$ is useless, solution works for any point on $A$-altitude. Let $F$ be the foot of $A$-altitude, we claim that $P$ lies on $(FQS)$. Let $H$ be the point on $(FQS)$ so that $FHQS$ is isosceles trapezoid. Let $K$ be the reflection of $B$ over $B$, let $B'$, $C'$ be the reflections of $B$, $C$ over $F$, respectively. By the homothety, $B'KEC$ is cyclic. Let $I$, $L$ be the reflections of $K,E$ over line $AF$. Then $BILC'$ is cyclic. Also note that $BCEI$ and $BCKL$ are parallelograms, since $BL\parallel CK$ and $BI\parallel CE$, and $L-I-K-E\parallel BC$. Now, $H$ is the midpoint of $CL$, $Q$ the midpoint of $CI$. Also, by radical axis on $(B'KEC)$, $(BECD)$ and $(BLIC')$ implies that $D$ lies on $(BLIC')$, all this is enough to prove that by the homothety $P$ lies on $(FHQS)$. Nice solution! The condition that $A'$ is the reflection of $A$ with respect to $BC$ is there to make the problem a bit easier and we had some solutions using it .

Here is a solution using trigonometry: Let $k$ be the circumcircle of $PQS$. Denote by $R$ the second intersection point of $k$ and $CD$. Then we have $\angle PRS = \angle PQS$. The line $PS$ is the midline in triangle $BCD$, so $PS \parallel BD$. Similarly, $QS$ is the midline in $CBE$, so $QS \parallel CE$ and $QS = \frac{1}{2} \cdot CE$. We also have that $\angle BDC = \angle BA'C = \angle BAC$, because of symmetry and the fact that $BDA'C$ is cyclic. From here we see that $CD = CA$ . We also get $\angle RPS = 180^{\circ} - \angle CPS = 180^{\circ} - \angle CDB = 180^{\circ} - \angle BAC$. From $QS \parallel AC$ and $PS \parallel AB$ and from orientation we have that $\angle PSQ = 180^{\circ} - \angle BAC$. Thus, $\angle RPS = \angle QSP$, $\angle PRS = \angle PQS$ and $PS = PS$ implies that the triangle $PRS$ is congruent to the triangle $SQP$ in that order, so $PR = SQ = \frac{1}{2} \cdot CE$. Also note that $AB = BE$, because $\angle BAE = \angle BA'C = \angle BDC = \angle BEA$ because of symmetry and the fact that $BDA'CE$ is a cyclic polygon. Now denote by $H$ the intersection point of $BC$ and $AA'$. We will show that $CS \cdot CH = CP \cdot CR$, which will imply that $H$ lies on the required circle by the converse of the power of a point theorem (or more elementary, just triangle similarity). We have the following calculations: $$CP \cdot CR = CP \cdot (CP + PR) = \frac{1}{2} \cdot CD \cdot (\frac{1}{2} \cdot CD + \frac{1}{2} \cdot CE)$$$$= \frac{1}{4} \cdot AC \cdot (AC + CE)$$Let us use the notation $AB = c, BC = a, CA = b$ and $\angle BAC = \alpha$, $\angle ABC = \beta$, $\angle ACB = \gamma$. Because $AB = BE$, we see that $AE = 2c \cos{\alpha}$. Hence, we have: $$CP \cdot CR = \frac{b(b+b-2c \cos{\alpha})}{4} = \frac{b(b-c \cos{\alpha})}{2}. $$It's easy to see that $CH = b \cos{ \gamma}$ and $CS = \frac{a}{2}$. Now we have: $$CP \cdot CR = CH \cdot CS \iff \frac{ab \cos{\gamma}}{2} = \frac{b(b-c \cos{\alpha)}}{2}$$$$\iff a \cos{ \gamma} = b - c \cos{\alpha} \iff a \cos{\gamma} + c \cos{\alpha} = b.$$The last equality is true because it is just the well known projection law.

rafaello wrote: The fact that $A'$ must be the reflection of $A$ wrt $BC$ is useless, solution works for any point on $A$-altitude. Let $F$ be the foot of $A$-altitude, we claim that $P$ lies on $(FQS)$. Let $H$ be the point on $(FQS)$ so that $FHQS$ is isosceles trapezoid. Let $K$ be the reflection of $B$ over $B$, let $B'$, $C'$ be the reflections of $B$, $C$ over $F$, respectively. By the homothety, $B'KEC$ is cyclic. Let $I$, $L$ be the reflections of $K,E$ over line $AF$. Then $BILC'$ is cyclic. Also note that $BCEI$ and $BCKL$ are parallelograms, since $BL\parallel CK$ and $BI\parallel CE$, and $L-I-K-E\parallel BC$. Now, $H$ is the midpoint of $CL$, $Q$ the midpoint of $CI$. Also, by radical axis on $(B'KEC)$, $(BECD)$ and $(BLIC')$ implies that $D$ lies on $(BLIC')$, all this is enough to prove that by the homothety $P$ lies on $(FHQS)$. Let $K$ be the reflection of $B$ over $B$ Can you explain it to me?

$AH$ Intersects $BC$ at $H$ and $SQ$ Intersects $AB$ at $G$ $SP$ Intersects $AC$ at $K$ Claim $1$:$HSKJ$ is cyclic Claim$2$:$HGQ$ and $HKP$ are Samilar $\angle{HGS}=\angle{HKS}$ $\frac{HG}{GQ}=\frac{AB}{AE}=\frac{AC}{AD}=\frac{HK}{SP}$ $\frac{AB}{AE}=\frac{AC}{AD}$ $\rightarrow$ $AEB$ and $ACD$ are Samilar

Nice problem! Let $AA'$ and $BC$ intersect at point $K$. Note that $QS \parallel EC$ and $SP \parallel BD$, and so $\angle QSP=180^\circ-\angle A$, hence it suffices to prove that $\angle QKP=90^\circ$. Let $B',C'$ be the reflections of $B,C$ across $K$. Then, $KQ \parallel B'E$ and $KP \parallel C'D$ and so $\angle QKP=180^\circ-\angle(C'D,B'E)$. Let lines $C'D$ and $B'E$ intersect at point $T$. Note that triangles $AC'D$ and $AB'E$ are similar, as $\angle C'AB=\angle B'AC$ and $AC'/AD=AC/AD=AB/AE=AB'/AE$. Therefore, we have $\angle(C'D,B'E)=\angle C'TE=360^\circ-\angle AC'D-\angle AEB-\angle C'AE=$ $=360^\circ-\angle AC'D-\angle ADC'-\angle C'AC=360^\circ-(180^\circ-\angle C'AB)-\angle C'AC=180^\circ-\angle A,$ as desired.