Note. The result still holds without the assumption that both triangles $ABP$ and $CDP$ are acute. This assumption helps the contestants to focus on more specific configurations. Indeed, because triangles are acute, $O_1$ lies inside triangle $ABP$ and $O_2$ lies inside triangle $CDP$. Points $M$ and $N$ lie in the region bounded by rays $PB$ and $PD$. It is not difficult to see that $O_1MNO_2$ is a convex quadrilateral. (In particular, this is helpful in solution 2.) Hence we can consider the configuration (in two diagrams) shown below. For other possible configurations, our proofs can be adjusted slightly.
We present two solutions. Both solutions are based on the fact that triangles $ABQ$ and $CDQ$ are similar isosceles triangles. Indeed, because $ABPQ$ and $DCPQ$ are cyclic, we have $\angle QBD = \angle QAC$.

and $\angle QDB = \angle QCA$. Hence triangles $ACQ$ and $BDQ$ are similar to each other. Because $AC = BD$, we conclude that triangle $ACQ$ and $BDQ$ are congruent to each other, implying that $QA = QB$ and $QC = QD$. Because $\angle BQD = \angle AQC$, we have $\angle AQB = \angle CQD$ and isosceles triangles $ABQ$ and $CDQ$ are similar to each other. In particular, $\angle QBA = \angle QAB = \angle QCD = \angle QDC$.

Solution 2. For point $X$ and line $\ell$, let $d(X, \ell)$ denote the distance from $X$ to $\ell$. Because $O_1MNO_2$ is a convex quadrilateral by the note prior to Solution 1, it suffices to show that
$$d(M, O_1O_2) = d(N, O_1O_2).$$
Working on arcs along $\omega_1$, we have
$$\begin{align*} \angle MO_1O_2 &= \angle MO_1P + \angle PO_1O_2 = \widehat{MP} + \frac{\angle PO_1Q}{2} = \frac{\widehat{SP}}{2} + \angle PBQ \\ &= \angle PBS + \angle PBQ = \angle SBQ = \angle CBQ. \end{align*}$$
Thus, we have
$$\frac{d(M, O_1O_2)}{O_1M} = \sin \angle MO_1O_2 = \sin \angle QBC = \frac{d(Q, BC)}{BQ} \quad \text{or} \quad \frac{d(M, O_1O_2)}{d(Q, BC)} = \frac{BQ}{O_1M}.$$
In exactly the same way, we can show that $\angle NO_2O_1 = \angle BCQ$ and
$$\frac{d(N, O_1O_2)}{d(Q, BC)} = \frac{CQ}{O_2N}.$$
It suffices to show that
$$\frac{BQ}{O_1M} = \frac{CQ}{O_2N},$$
which is holds because $BQ$ and $CQ$ are two corresponding sides of two similar (isosceles) triangles (namely, $BAQ$ and $CDQ$) inscribed in circles $\omega_1$ and $\omega_2$, with radii $O_1M$ and $O_N$, respectively.