We will use inversion at $E$ with power $-EB \cdot ED.$ Let $X'$ denote the image of a point $X$, so that $C = A', D = B'.$ Let $\omega_1 = (\triangle ECD), \omega_2 = (\triangle EAD).$ Observe that the images of lines $BC, AB$ under this inversion are $\omega_2, \omega_1$ respectively. Now, note that $P' = \ell \cap \omega_1, Q' = \ell \cap \omega_2,$ and $R'$ is the point on $(O)$ so that $BR' || \ell.$ It suffices to show that $DQ'R'P'$ is cyclic because $D = B'.$ Observe that $\angle DCR' \angle DBR' = \angle DEQ'$ and $\angle CR'D = 180 - \angle CAD = 180 - \angle EAD = \angle EQ'D.$ This means that $\triangle EQ'D \sim \triangle CR'D.$ Therefore, there exists a spiral similarity sending $\triangle DQ'E$ to $\triangle DR'C$ centered at $D$, and so hence we've $\triangle DQ'R' \sim \triangle DEC.$ Finally, we have that $\angle DR'Q' = \angle DCE$ (from the similarity), and $\angle DCE = \angle DP'E = \angle DP'Q'$. Hence, as $\angle DR'Q' = \angle DP'Q'$ we have that $DQ'R'P'$ is cyclic and we're done. $\square$

Rewrite R is Miquel point of ABCPQE so RDE=RAP=REP hence (RED) tangent PQ q.e.d

https://artofproblemsolving.com/community/c6h1622451p10157861

We have $\angle{RAB}=\angle{RDE}=\angle{REP}$ so $ARPE$ is cyclic. Hence $\angle{RPQ}=\angle{RAE}=\angle{RBQ}$. Hence $RPBQ$ is cyclic.

Denote $\measuredangle$ as the directed angle modulo $180^{\circ}$ Now, Claim 01. $PARE$ is cyclic. Proof. Notice that $\measuredangle PER = \measuredangle EDR \equiv \measuredangle BDR = \measuredangle BAR \equiv \measuredangle PAR$. Claim 02. $ERQC$ is cyclic. Proof. Notice that $\measuredangle REQ = \measuredangle REP = \measuredangle RCB = \measuredangle RCQ$. Now, we have $\measuredangle ARP = \measuredangle AEP = \measuredangle CEQ = \measuredangle CRQ$. Since $\measuredangle PAR \equiv \measuredangle BAR = \measuredangle QCR$. So by spiral similarity, $\measuredangle ARC = \measuredangle PRQ$. But we know $\measuredangle PRQ = \measuredangle ARC = \measuredangle ABC = \measuredangle PBQ$, as desired.

$\angle PER=\angle BDR=\angle BAR=\angle PAR$. So $ARPE$ is cyclic. $\angle QER=\angle PER=\angle BAR=\angle BCR=\angle QCR$. So $QREC$ is cyclic too. $\angle RQB=\angle RQC=180-\angle REC=\angle AER=\angle APR$.

Let $X=RE\cap (ABCD)$, $Y=RP\cap (ABCD)$. By Pascal's theorem on $YRXABD$ we find that $P-E-XA\cap YD$ are colinear. By Pascal on $BXACYD$ we find that $\infty_{BX}-XA\cap YD-E$ are also colinear. This implies that $PE\parallel BX$ and thus by Reim's or whatever $PEAR$ is cyclic. Hence $\triangle RBP\sim \triangle RCE$ and thus we conclude the problem by properties of sprial similarity. Hence we win

Claim: $PARE$ is cyclic. Proof. $\measuredangle REP=\measuredangle RDE=\measuredangle RDB=\measuredangle RAB=\measuredangle RAP.$ $\blacksquare$ Thus, $\measuredangle QBR=\measuredangle CBR=\measuredangle CAR=\measuredangle EAR=\measuredangle EPR=\measuredangle QPR.$ $\square$

Let $R'$ be the Miquel point of complete quadrilateral $QBCEAP$. Note that $$\measuredangle R'EP=\measuredangle R'AP=\measuredangle R'AB=\measuredangle R'DB=\measuredangle R'DE,$$thus $R\equiv R'$. Hence, we conclude that $BQRP$ is cyclic.

By chasing angles: $\angle BDR=\angle RAP=\angle QER=\angle QCR$ In the $ABCEQP$, $R \in \odot (ECR), \odot (AEP) \implies R=\text{Miquel Point}$ $\implies B,P,R,Q$ are concyclic.