Let $K = PQ \cap (BPE)$, and bisector of $DQ$ meet $PD, DK$ at $L, T$. $PQ \cdot QK = BQ \cdot QE = AQ \cdot QD \implies P, A, K, D$ concyclic. $PA = AK \implies \angle PDQ = \angle KDQ$ $QL = LD = DT = TQ \implies QLDT$ is rhombus $\implies QL \parallel DK$ $CD \perp DQ \text{ \ and \ } LR \perp DQ \implies CD \parallel LR$ $\frac{BS}{SP} = \frac{CR}{RP} = \frac{DL}{LP} = \frac{KQ}{QP} \implies BK \parallel SQ \implies (BPE), (PQS)$ are tangent. $\blacksquare$

Here is a ratio-chasing solution that doesn't require a brain. It's not hard to see that the conclusion is equivalent to $\angle SQP=\angle BEP$ (and this implies that $\angle BQS=\angle EPQ$). Now, by ratio lemma in $\triangle BQP$ and LOS in $\triangle PEQ$, this is equivalent to \[\frac{QP}{QE}=\frac{PS}{SB}\cdot\frac{BQ}{QP}\iff \frac{QP^2}{EQ\cdot QB}=\frac{PS}{SB}\]By PoP and Thales, this is further equivalent to \[\frac{QP^2}{AQ\cdot QD}=\frac{PR}{RC}\]Now let $f(x)=XQ^2-XD^2$ note that this is linear so \[\frac{PR}{RC}=\frac{f(P)-f(R)}{f(R)-f(C)}=\frac{PD^2-PQ^2}{CQ^2-CD^2}=\frac{PD^2-PQ^2}{DQ^2}\]The last equality comes from $\angle QDC=90^\circ$. Now by shooting lemma, we have that $AP^2=AQ\cdot AD$, or $AP$ is tangent to $(DPQ)$. Finally, we can rewrite the conclusion as \[\frac{DQ}{AQ}=\frac{PD^2}{PQ^2}-1\]It is well known that $\frac{PD^2}{PQ^2}=\frac{AD}{AQ}$ (for example by ratio lemma) so the conclusion follows. $\blacksquare$