, 2022

Let $D', E'$ be points on lines $AB, AC$, respectively, such that $D'E'$ is parallel to $DE$, and passes through $N$. From
$$\angle ND'B = 90^\circ - \angle BAQ = \angle OQB = \angle NQB,$$
we see that $B, D', N, Q$ are concyclic. Since
$$\angle QD'E' = \angle QBC = \angle QAE',$$
$A, D', Q, E'$ are concyclic. Therefore, $\triangle ADE \cup P \stackrel{+}{\sim} \triangle AD'E' \cup Q$.
Let $S', T \neq A$ be the intersections of $AS$ and $D'E'$, $\Omega$, respectively. Then
$$\angle QTS' = \angle QTA = \angle(QO, \perp AQ) = \angle QNS'$$
gives the concyclicity of $Q, T, N, S'$. From the similarity $\triangle ADE \cup \{P, S\} \stackrel{+}{\sim} \triangle AD'E' \cup \{Q, S'\}$, we see that $PS$ is parallel to $QS'$. According to Reim's theorem, $S, N, O, T$ are concyclic. Thus,
$$\angle OWS = \angle SAO = \angle OTA = \angle ONS,$$
as desired.