Let $S$ be a point such that $PESF$ is a parallelogram (Fig. 15). We first show that $PR \parallel AS$. For this note that $\angle PER = \angle SFA$ since $PE \parallel FS$ and $ER \parallel AF$. Additionally $PE = FS$ and $ER = AF$, thus triangles $PER$ and $SFA$ are congruent. From this we deduce that $PR \parallel AS$.

In order to prove the problem statement it now suffices to show that $A$, $S$ and $Q$ are collinear. Let $X$ be the point of intersection of $ES$ and $AF$ and $Y$ be the point of intersection of $FS$ and $AE$ (Fig. 16).

First we show that $XY \parallel BC$. Denote $\frac{AE}{AC} = \kappa$ and $\frac{AF}{AB} = \lambda$. As $EX \parallel CF$, we have $\frac{AX}{AF} = \frac{AE}{AC} = \kappa$. Thus $\frac{AX}{AB} = \frac{AX}{AF} \cdot \frac{AF}{AB} = \kappa \cdot \lambda$. Analogously, we get $\frac{AY}{AC} = \kappa \cdot \lambda$. Consequently, $\frac{AX}{AB} = \frac{AY}{AC}$, which implies $XY \parallel BC$.

The fact just proven implies $\frac{XY}{BC} = \frac{AX}{AB}$. It suffices to notice that the triangles $BCQ$ and $XYS$ are similar as their respective sides are parallel. Thus $\overrightarrow{XS} = \overrightarrow{XY}$ which implies $\overrightarrow{XS} = \frac{AX}{AB}$. Since $XS \parallel BQ$, this implies that the points $A$, $S$ and $Q$ are collinear.

Fig. 15

Fig. 16

Denote $\vec{AC} = \vec{e}_1$ and $\vec{AB} = \vec{e}_2$. Take $\kappa, \lambda$ such that $\vec{AE} = \kappa \vec{e}_1$ and $\vec{AF} = \lambda \vec{e}_2$. Then
$$\begin{aligned} \vec{AR} &= \vec{AE} + \vec{AF} = \kappa \vec{e}_1 + \lambda \vec{e}_2, \\ \vec{EB} &= \vec{AB} - \vec{AE} = \vec{e}_2 - \kappa \vec{e}_1, \\ \vec{FC} &= \vec{AC} - \vec{AF} = \vec{e}_1 - \lambda \vec{e}_2. \end{aligned}$$
Take also $\alpha, \beta$ such that $\vec{EP} = \alpha \vec{EB}$ and $\vec{FP} = \beta \vec{FC}$. Then
$$\begin{aligned} \vec{AP} &= \vec{AF} + \beta \vec{FC} = \lambda \vec{e}_2 + \beta(\vec{e}_1 - \lambda \vec{e}_2) = \beta \vec{e}_1 + (1 - \beta)\lambda \vec{e}_2, \\ \vec{AP} &= \vec{AE} + \alpha \vec{EB} = \kappa \vec{e}_1 + \alpha(\vec{e}_2 - \kappa \vec{e}_1) = (1 - \alpha)\kappa \vec{e}_1 + \alpha \vec{e}_2. \end{aligned}$$
Consequently, $\beta \vec{e}_1 + (1 - \beta) \lambda \vec{e}_2 = (1 - \alpha) \kappa \vec{e}_1 + \alpha \vec{e}_2$. As vectors $\vec{e}_1$ and $\vec{e}_2$ are not parallel, the equality can hold only if the respective coefficients are equal, i.e., $\beta = (1 - \alpha)\kappa$ and $(1 - \beta)\lambda = \alpha$. This implies $\vec{AP} = \beta \vec{e}_1 + \alpha \vec{e}_2$. Now we obtain
$$\begin{aligned} \vec{PR} &= \vec{AR} - \vec{AP} = (\kappa \vec{e}_1 + \lambda \vec{e}_2) - (\beta \vec{e}_1 + \alpha \vec{e}_2) = (\kappa - \beta) \vec{e}_1 + (\lambda - \alpha) \vec{e}_2, \\ \vec{AQ} &= \vec{AC} + \vec{CQ} = \vec{AC} + \vec{PB} = \vec{AC} + \vec{AB} - \vec{AP} \\ &= \vec{e}_1 + \vec{e}_2 - (\beta \vec{e}_1 + \alpha \vec{e}_2) = (1 - \beta) \vec{e}_1 + (1 - \alpha) \vec{e}_2. \end{aligned}$$
The lines $PR$ and $AQ$ being parallel is equivalent to the vectors $\vec{PR}$ and $\vec{AQ}$ being parallel, which in turn is equivalent to their coefficients being proportional. Hence it suffices to show that $\frac{\kappa - \beta}{1 - \beta} = \frac{\lambda - \alpha}{1 - \alpha}$, or equivalently,
$$(\kappa - \beta)(1 - \alpha) = (\lambda - \alpha)(1 - \beta).$$
As $\kappa(1 - \alpha) = \beta$ and $\lambda(1 - \beta) = \alpha$, we have
$$\begin{aligned} (\kappa - \beta)(1 - \alpha) &= \kappa(1 - \alpha) - \beta(1 - \alpha) = \beta - \beta + \alpha\beta = \alpha\beta, \\ (\lambda - \alpha)(1 - \beta) &= \lambda(1 - \beta) - \alpha(1 - \beta) = \alpha - \alpha + \alpha\beta = \alpha\beta. \end{aligned}$$
Hence the desired equality holds and the proof is complete.