A pretty enjoyable problem. I hope I didn't trip on this its not very difficult. Claim : $FH \parallel AB$. Proof : We let $H'$ be the intersection of the line through $F$ parallel to $AB$ and $AD$. Let $G' =\overline{BH'} \cap \overline{CE}$. Then, note that, $\measuredangle BFH' = 90^\circ = \measuredangle EBC$. Further, $BF = BE$ and $FH' = AB=BC$. Thus, $\triangle BFH' \cong \triangle EBC$. Now, this means that, \[\measuredangle EBG' = \measuredangle EBH' = \measuredangle FH'B = \measuredangle BCE = \measuredangle BCG'\]from which it is clear that $BG' \perp CE$. This implies that $G'=G$ and $H'=H$ which proves the desired claim. Claim : Quadrilateral $BGPF$ and pentagon $HGFCD$ are cyclic. Proof : Note that from the above claim, it is clear that \[\measuredangle BFP = \measuredangle BFH = 90^\circ = \measuredangle CGB = \measuredangle PGB\]from which it follows that $BGPF$ is cyclic as desired. Further, \[\measuredangle HGC = 90^\circ = \measuredangle HFC = 90^\circ = \measuredangle HDC\]which proves that the points $H$ , $G$ , $F$ , $C$ and $D$ are concyclic as claimed. Now, note that by Radical Center Theorem on circles $(GCD)$ , $(GBF)$ and $(CBD)$ $Q$ must lie on the radical axis of circles $(GBF)$ and $(CBD)$, as it lies on the other two radical axes according to its definition. Thus, if $R = \overline{DP} \cap (BGF)$. It follows that $R$ also lies on $(BCD)$ since \[\measuredangle BRD = \measuredangle BRP = \measuredangle BGP =90^\circ\]Thus, points $B$ , $R$ and $Q$ must be collinear which implies that $R= \overline{BQ} \cap \overline{DP}$ which proves the desired statement.

Pretty easy to coordbash, but here's a synthetic sol since I don't like mindlessly bashing. Let $\omega$ be the circumcircle of $ABCD$, and $\gamma$ the circumcircle of $BGPF$ which exists because $\angle BGP = \angle BFP = 90$. Let $K$ be the intersection of $\omega$ and $\gamma$ other than $B$. Claim 1: $BFG$ and $CDG$ are spirally similar about point $G$. We show that these triangles are similar with scale factor $\frac{BE}{AB}$. Because $BGC \sim EBC$, we see that $\frac{BG}{CG} = \frac{BE}{AB}$. Additionally, $\frac{BF}{CD} = \frac{BE}{AB}$. Finally, $\angle GBF = 90 - \angle EBG = 90 - \angle BCE = \angle BCG$, so the claim is complete. Claim 2: $Q$ lies on the radical axis of $\omega$ and $\gamma$. Observe that $\angle BGF = \angle BGC + \angle CGF = \angle CGF + \angle FGB = 90$ by Claim 1. Therefore, $CDGF$ is cyclic, so $QC \cdot QD = QF \cdot QG$, and $Q$ is the intersection of $CD$ and $FG$, so the power of $Q$ is the same for both circles. Claim 3: $BQ \perp DP$. Observe that $B$, $K$, and $Q$ are collinear by Claim 2. Additionally, $\angle PKB = \angle DKB = 90$ because $BP$ and $BD$ are the diameters of $\gamma$ and $\omega$, respectively. This implies $D$, $P$, $K$ are collinear and $\angle DKB = 90$, hence $\angle DKQ = 90$, and we are done.

Here is an elementary coordinate bashing .I will say it aint beautiful but that was lots of fun calculations Let the origin $D(0,0)$ be one vertex of the square with side length $\ell$, such that $AD$ and $CD$ lie on the $Y$-axis and $X$-axis, respectively. Let $BE = BF = r$. The equation of $EC$ in two-point form is: \[ \ell x + ry = \ell^2. \] Since $BH \perp EC$, the equation of $BH$ is: \[ rx - \ell y = k. \]Plugging in point $B(\ell, \ell)$ (as $B$ lies on $BH$), we get: \[ r\ell - \ell^2 = k. \]Thus, the equation of $BH$ becomes: \[ rx - \ell y = r\ell - \ell^2. \]Putting $x = 0$, the $y$-coordinate of $H$ is: \[ y = \ell - r. \]Thus, $H=(0, \ell - r)$, and we conclude that $AB \parallel HF$. Solving the equations of $EC$ and $BH$, we find the coordinates of $G$: \[ G\left( \frac{\ell^3 + r^2 - \ell^2r}{\ell^2 + r^2}, \frac{\ell^3}{\ell^2 + r^2} \right). \] The equation of $HF$ is: \[ y = \ell - r. \] Solving the equations of $HF$ and $EC$, we find the coordinates of $P$: \[ P\left( \frac{\ell^2 - r^2 - \ell r}{\ell}, \ell - r \right). \] The slope of $PD$ is: \[ m_1 = \frac{\ell^2 - \ell r}{\ell^2 + r^2 - \ell r}. \] The equation of $GF$ in two-point form is: \[ y - (\ell - r) = \frac{\ell^2 - r\ell + r^2}{-\ell^2}(x - \ell). \]Putting $y = 0$, we find the coordinates of $Q$: \[ Q\left( \frac{2\ell^3 - 2r\ell^2 - r^2\ell}{\ell^2 + r^2 - r\ell}, 0 \right). \] The slope of $BQ$ is: \[ m_2 = \frac{r\ell^2 - \ell^3 - r^2\ell}{\ell^3 - r\ell^2}. \] Finally, we calculate: \[ m_1 \cdot m_2 = \frac{\ell^2 - \ell r}{\ell^2 + r^2 - \ell r} \cdot \frac{-\ell(\ell^2 + r^2 - \ell r)}{\ell(\ell^2 - \ell r)} = -1. \] Thus, $DP \perp BQ$.