We will prove that the lines $\overline{AB}$, $\overline{CD}$, and $\overline{Q_1Q_2}$ are either concurrent or all parallel. Let $X$ and $Y$ denote the reflections of $P$ across the lines $\overline{AB}$ and $\overline{CD}$.

We first claim that $XQ_1 = YQ_1$ and $XQ_2 = YQ_2$. Indeed, let $Z$ be the reflection of $Q_1$ across $BC$. Then $XB = PB$, $BQ_1 = BZ$, and
$$\angle XBQ_1 = \angle XBA + \angle ABQ_1 = \angle ABC = \angle PBC + \angle CBZ = \angle PBZ,$$
whence $\triangle XBQ_1 \cong \triangle PBZ$ and thus $XQ_1 = PZ$. Similarly $YQ_1 = PZ$, and so $XQ_1 = YQ_1$. In exactly the same way, we see that $XQ_2 = YQ_2$, establishing the claim. We conclude that line $\overline{Q_1Q_2}$ is the perpendicular bisector of the segment $\overline{XY}$.

Now, if $\overline{AB} \parallel \overline{CD}$, then $\overline{XY} \perp \overline{AB}$ and it follows that $\overline{Q_1Q_2} \parallel \overline{AB}$, as desired. If lines $\overline{AB}$ and $\overline{CD}$ are not parallel, then let $R$ denote their intersection. Since $RX = RP = RY$, $R$ lies on the perpendicular bisector of $\overline{XY}$ and thus $R, Q_1$, and $Q_2$ are collinear, as desired.

We approach the problem using isogonal conjugates. Recall that two points $S$ and $T$ are isogonal conjugates with respect to $ABC$ if $\angle SAB = \angle CAT$, $\angle SBC = \angle ABT$, and $\angle SCA = \angle BCT$, with any two of these equalities implying the third.

If $\overline{AB} \parallel \overline{CD}$, then there is nothing to prove; thus we assume $\overline{AB}$ intersects $\overline{CD}$ in a point $R$. Then $Q_1$ and $P$ are isogonal conjugates with respect to $RBC$, whence $\angle Q_1RB = \angle CRP$. Similarly, $Q_2$ and $P$ are isogonal conjugates with respect to $RAD$, whence $\angle Q_2RA = \angle DRP$. Therefore $\angle Q_1RB = \angle Q_2RA = \angle Q_2RB$ and the lines $\overline{AB}$, $\overline{CD}$, $\overline{Q_1Q_2}$ all intersect at $R$.