Let $ M $ be the midpoint of arc $ AD $ and $ S^* $ $ \equiv $ $ QM $ $ \cap $ $ \omega $. From $ QE $ $ \cdot $ $ QF $ $ = $ $ QA $ $ \cdot $ $ QP $ $ = $ $ QC $ $ \cdot $ $ QR $ $ = $ $ QM $ $ \cdot $ $ QS^* $ $ \Longrightarrow $ $ C, $ $ E, $ $ F, $ $ R $ are concyclic and $ E, $ $ F, $ $ M, $ $ S^* $ are concyclic, so notice that $ DM $ $ \parallel $ $ EF $ we get $ E, $ $ M, $ $ R $ are collinear (Reim's theorem) $ \Longrightarrow $ $ F $ is the Miquel point of $ \triangle MQR $ with the points $ C $ $ \in $ $ QR, $ $ E $ $ \in $ $ RM, $ $ S^* $ $ \in $ $ MQ $, hence $ C, $ $ F, $ $ Q, $ $ S^* $ are concyclic. Since $ \measuredangle S^*QE $ $ = $ $ \measuredangle S^*CD $ $ = $ $ \measuredangle S^*BE $, so $ B, $ $ E, $ $ Q, $ $ S^* $ are concyclic $ \Longrightarrow $ $ S^* $ $ \equiv $ $ S $. Since $ \measuredangle RBE $ $ = $ $ \measuredangle RCF $ $ = $ $ \measuredangle REQ $, so combine $ \measuredangle ERB $ $ = $ $ \measuredangle QRE $ we get $ \triangle BER $ $ \sim $ $ \triangle EQR $, hence $ \measuredangle BER $ $ = $ $ \measuredangle EQR $ $ = $ $ \measuredangle FQC $ $ = $ $ \measuredangle FSC $.

Since \[\measuredangle FCS = \measuredangle DCS = \measuredangle DBS = \measuredangle EBS = \measuredangle EQS = \measuredangle FQS,\]it follows that $F, S, C, Q$ are concyclic. Therefore, $\measuredangle FSC = \measuredangle FQC.$ By Power of a Point, $QE \cdot QF = QA \cdot QP = QR \cdot QC \implies E, F, R, C$ are concyclic. Therefore, $\measuredangle REF = \measuredangle QCF.$ Meanwhile, as $DE = DF$, we have $\measuredangle BEF = \measuredangle DEF = \measuredangle EFD = \measuredangle QFC.$ It follows that \[\measuredangle BER = \measuredangle BEF - \measuredangle REF = \measuredangle QFC - \measuredangle QCF = \measuredangle FQC. \; \square\]

Darn I should draw better diagrams at tests. Note that $A, E, F, P$ lie on a circle, so by PoP, we have $QR \cdot QC = QP \cdot QA = QF \cdot QE$, so $R,C,F,E$ is cyclic. Now $\angle FCS = \angle DCS = \angle DBS =\angle FQS$, so $C, Q, F, S$ is cyclic. We now have $$\angle BER = 180-\angle REF - \angle FED = 180-\angle QCF - \angle EFD = 180-\angle QCF - \angle QFC = \angle CFQ = \angle CQF = \angle CSF$$as desired.

First,as $QF*QE=QP*QA=QC*QR$,we get $EFRC$ are concylic; Next,as $\angle ERQ = \angle DFE = \angle DEF$,we get $(EQR)$ is tangent to line $DB$, Hence we just need to prove that $SFCQ$ is concylic; which is obvious,as $\angle SQF = \angle SQE = \angle SBE =\angle SBD =\angle SCD$.KO!

Notice that $Q=\overline{RC}\cap\overline{EF}\cap\overline{AP},$ so $CEFR$ is cyclic by Radical Axis. Also, $$\measuredangle DCS=\measuredangle DBS=\measuredangle DES=\measuredangle EQS=\measuredangle FQS$$so $CQSF$ is cyclic. Hence, \begin{align*}\measuredangle CSF&=\measuredangle CQF\\&=\measuredangle CRE+\measuredangle REF\\&=\measuredangle CFE+\measuredangle REF\\&=\measuredangle DFE+\measuredangle REF\\&=\measuredangle FED+\measuredangle REF\\&=\measuredangle FEB+\measuredangle REF\\&=\measuredangle REB.\end{align*}$\square$