Let $G$, $X$, and $Y$ be the respective feet of the perpendiculars from $P$ to $EF$, $EC$, and $FB$. Note that $EPYB$ and $FPXC$ are cyclic quadrilaterals, so
$$\begin{aligned} \angle EYF &= 90^\circ + \angle EYP = 90^\circ + \angle EBP = 90^\circ + \angle ABP \\ &= 90^\circ + \angle DCP = 90^\circ + \angle FCP = 90^\circ + \angle FXP = \angle FXE. \end{aligned}$$
Thus, $EXYF$ is a cyclic quadrilateral. Note that $EGPX$ and $FGPY$ are also cyclic. The radical axes of the circumcircles of $EXYF$, $EGPX$, and $FGPY$ with each other are $GP$, $EX$, and $FY$. Thus, by the radical axis theorem, $GP$, $EX$, and $FY$ concur. Since $Q$ is the intersection of $EX$ and $FY$, by the construction of $G$ we have $GP \perp EF$, so it follows that $QP \perp EF$.

We adopt the notations of the previous solution. Define point $R_E$ on $PG$ so that $ER_E \perp BF$. Because $\angle GR_EX = \angle GFX = 90^\circ$, quadrilateral $XRE_EFG$ is cyclic. Hence, we have
$$\angle PR_E E = \angle PR_E X = \angle GR_E X = \angle GFX = \angle EFB. \qquad (42)$$
Also note that by the definitions of $E$ and $R_E$, we have
$$\angle R_E EP = 90^\circ - \angle BER_E = 90^\circ - \angle BEX = \angle XBE = \angle FBE. \qquad (43)$$
By (42) and (43), we know that $\triangle PR_E E \sim \triangle EFB$, implying that $\frac{PR_E}{EF} = \frac{EP}{BE}$. Defining $R_F$ to be the point on $PG$ such that $FR_F \perp CE$, we see in a similar manner that $\frac{PR_F}{EF} = \frac{FD}{CF}$. Finally, triangles $ABP$ and $DCP$ in cyclic quadrilateral $ABCD$ are similar with $E$ and $F$ being corresponding points, we have $\frac{EP}{BE} = \frac{FP}{CF}$. Combining these equalities of ratios, we find that
$$\frac{PR_E}{EF} = \frac{EP}{BE} = \frac{FP}{CF} = \frac{PR_F}{EF},$$
hence the points $R_E$ and $R_F$ are the same point, which we call $R$. We now see that $Q$ is the orthocenter of triangle $EFR$, so in particular $RQ \perp EF$. On the other hand, we have $RP \perp EF$ by definition, meaning that $R, Q$, and $P$ are collinear and $PQ \perp EF$.

Let $M$ and $N$ be the midpoints of $BC$ and $AD$, respectively. We begin with a lemma.

Lemma 3 (Kvant 2007). Quadrilateral MFNE is a kite, meaning that MN $\perp$ $EF$ and MN passes through the midpoint of $EF$.
Proof. Let $K$ and $L$ be the midpoints of $AP$ and $DP$. We have that $EK = \frac{1}{2}AP = LN$ and $KN = \frac{1}{2}DP = FL$; further, because $NLPK$ is a parallelogram and $\angle PKE = 2\angle PAE = 2\angle FDP = \angle FLP$, we see that $\angle NKE = \angle FLN$. Together, these show that $\triangle EKN \simeq \triangle NLF$, hence $NE = NF$. Similarly, we obtain $ME = MF$, which yields the desired result. $\square$

By Lemma 3, it suffices for us to prove that $PQ \parallel MN$. If $AB \parallel CD$, this is clear. Otherwise, define $Z$ to be the intersection of $AB$ and $CD$. Letting $U$ be the midpoint of $PZ$, by Lemma 2 applied to $ACDB$, $U$ lies on line $MN$. Further, by Lemma 3, the midpoint $W$ of $EF$ also lies on this line.
Now, let $S$ be the intersection of $AF$ and $DE$. By Pappus' theorem on points ($A$, $E$, $B$) and ($D$, $F$, $C$), we find that $S$ lies on $PQ$. By Lemma 2 on $AEDF$, we see that the midpoint $V$ of $SZ$ lies on $NW$, hence on $MN$. In particular, the lines $MN$ and $UV$ coincide.

Now consider the homothety about $K$ with ratio $\frac{1}{2}$. It sends $P$ to $U$ and $S$ to $V$, hence it sends line $PS$ to $UV$. But $PQ$ and $PS$ coincide and $UV$ and $MN$ coincide, so this shows that $PQ \parallel MN$.