Actually a very nice problem.

Let $R$ be the intersection of lines $AE$ and $BC$. Point $Q$ lies on $(CRE)$ due to $\angle CQE = \angle CBA + \angle BAE = \angle CRE$. Clearly $R$ is the radical centre of circles $(PAE), (CBP),$ and $(ABCDE)$, hence $R$ lies on $PQ$. Now $R$ is also the radical centre of circles $(DPQ)$, $(PAE)$ and $(ABCDE)$, so the second intersection $K$ of circles $(DPQ)$ and $(ABCDE)$ lies on $DR$. Since $DR$ and $(ABCDE)$ are fixed, so is $K$. The circumcentre then lies on the bisector of $DK$, which is fixed.

The internal angles of this pentagon are all $72^{\circ}$. Let $T$ be the intersection of lines $AE$ and $BC$. Let $\omega$ be the circumcircle of triangle $TEC$. We claim that $Q\in\omega$. Note that $\angle CTE=36^{\circ}$ and $$\angle EQP=\angle EAP=72^{\circ}=\angle PBC=\angle PQC\Longrightarrow\angle EQC=144^{\circ}=\angle ETC\Longrightarrow Q\in\omega.$$Note that since $TA\cdot TE=TB\cdot TC$, point $T$ has same power w.r.t circles $(PAE)$ and $(PBC)$. So, $T,P,Q$ are collinear. Moreover, $TP\cdot TQ=TA\cdot TB$. Let $F$ be the center of $\omega$. Then,$$\angle CFE=2\angle CTE=72^{\circ}=\angle CDE.$$So, $F\in(CDE)$. Since $ABCDE$ is regular, it is cyclic as well. So, $F\in(ABCDE)$. Moreover, points $T,F,D$ all lie on the perpendicular bisector of $AB$. So, $$TF\cdot TD=TA\cdot TE=TP\cdot TQ.$$We conclude that $F\in(DPQ)$. So, $R$ lies on the perpendicular bisector of $FD$, which is the required (fixed) line.

Let $M$ be the midpoint of small arc $AB$. Then note that $DPQM$ is cyclic so the circumcenter of $\triangle DPQ$ is just the circumcenter of $\triangle DPM$ which clearly lies on the perpendicular bisector of $DM$.

I claim the fixed line is the one passing through the center of $(ABCDE)$ parallel to $\overline{AB}$. Let $X \neq D:=(DPQ) \cap (ABCDE)$. First, by radical center on $(APE)$, $(BCP)$, and $(ABCDE)$, we find that $\overline{AE}, \overline{BC}, \overline{PQ}$ concur. Then, by radical center on $(APE), (DPQ), (ABCDE)$, we find that $\overline{AE}, \overline{PQ}, \overline{DX}$ concur, so these two concurrency points must be identical. Then by symmetry it follows that $X$ is the point on $(ABCDE)$ diametrically opposite $D$, and the conclusion follows. $\blacksquare$ Remark: $P$ lying on $\overline{AB}$ is completely irrelevant lol

Let $X=\overline{AE}\cap\overline{BC}$, $\omega_B=(BCQP)$, $\omega_E=(AEQP)$, $\omega=(ABCDE)$, and $\gamma=(DPQ)$. Note by radical axis on $\omega_E$, $\omega_B$, and $\omega$ we have $X\in\overline{PQ}$. Also, radical axis on $\gamma$, $\omega$, $\omega_B$ yields $X\in\overline{DY}$ where $Y=\omega\cap\gamma$. Hence, $R$ lies on the perpendicular bisector of $\overline{DY}$, which is fixed. $\square$

Solution from Twitch Solves ISL: In fact, the condition that $ABCDE$ is regular is needlessly strong. The conclusion holds for any cyclic pentagon. [asy][asy] size(12cm); pair M = dir(260); pair D = dir(90); pair A = dir(240); pair E = dir(170); pair B = dir(300); pair T = extension(E, A, M, D); pair O = origin; pair C = -B+2*foot(O, T, B); draw(unitcircle); filldraw(A--B--C--D--E--cycle, invisible, blue); filldraw(unitcircle, invisible, blue); draw(A--T--B, blue); draw(D--T, heavycyan); pair P = 0.65*B+0.35*A; pair Q = -P+2*foot(circumcenter(E, A, P), P, T); draw(circumcircle(E, A, P), deepgreen); draw(circumcircle(B, C, P), deepgreen); filldraw(circumcircle(D, P, Q), invisible, red); pair R = circumcenter(D, P, Q); draw(T--Q, dashed); draw(R--O, lightred+dashed); dot("$M$", M, dir(315)); dot("$D$", D, dir(D)); dot("$A$", A, dir(A)); dot("$E$", E, dir(E)); dot("$B$", B, dir(B)); dot("$T$", T, dir(T)); dot("$O$", O, dir(0)); dot("$C$", C, dir(C)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ !size(12cm); M 315 = dir 260 D = dir 90 A = dir 240 E = dir 170 B = dir 300 T = extension E A M D O 0 = origin C = -B+2*foot O T B unitcircle A--B--C--D--E--cycle / 0.1 cyan / blue unitcircle / 0.1 lightcyan / blue A--T--B / blue D--T / heavycyan P = 0.65*B+0.35*A Q = -P+2*foot (circumcenter E A P) P T circumcircle E A P / deepgreen circumcircle B C P / deepgreen circumcircle D P Q / 0.1 palered / red R = circumcenter D P Q T--Q / dashed R--O / lightred dashed */ [/asy][/asy] Claim: $D$, $P$, $Q$, $M$ are cyclic. Proof. Let $T \coloneqq \overline{AE} \cap \overline{BC}$ and let $\overline{DT}$ meet the circumcircle of the pentagon again at $M$. Since $TA \cdot TE = TB \cdot TC$, it follows $T$ lies on the radical axis $\overline{PQ}$. Then $TP \cdot TQ = TM \cdot TD$. $\blacksquare$ Hence $R$ lies on the perpendicular bisector of $\overline{DM}$, which does not depend on the choice of $P$ and is thus the desired fixed line.

Let $X$ be the intersection of lines $EA$ and $BC$, and let $M$ be the midpoint of minor arc $AB$ on $(ABCDE)$. Since $XA \cdot XE = XB \cdot XC$, we have that $X$ lies on the radical axis of $(AEPQ)$ and $(BCPQ)$, which is line $PQ$. So, \[XP \dot XQ = XA \cdot XE = XM \cdot XD,\]which implies $PQDM$ is cyclic. Therefore, the center of $(PQDM)$ always lies on the perpendicular bisector of line $DM$.

Let $X$ be the intersection of lines $AE$ and $BC$, and let $M$ be the intersection of circles of $ABCDE$ and $PQD $ Observe $ XA.XE = XP.XQ = XM.XD$. Since$ D $ is fixed it means that $M$ is fixed.( It is the midpoint of $arc$ $ AB$). Which implies that the radical axis is fixed which implies that $R$ varies as a straight line since center of$ ABCDE $ is fixed