Let $S$ be the tangent point of the circles $O$ and $O_{1}$ and let $T$ be the intersection point, different from $S$, of the circle $O$ and the line $SP$. Let $X$ be the tangent point of $\ell$ to $O_{1}$ and let $M$ be the midpoint of the line segment $XP$. Since $\angle TBP = \angle ASP$, the triangle $TBP$ is similar to the triangle $ASP$. Therefore,
$$\frac{PT}{PB} = \frac{PA}{PS}$$
Since the line $\ell$ is tangent to the circle $O_{1}$ at $X$, we have
$$\angle SPX = 90^\circ - \angle XSP = 90^\circ - \angle PAM = \angle PAM$$
which implies that the triangle $PAM$ is similar to the triangle $SPX$. Consequently,
$$\frac{XS}{XP} = \frac{MP}{MA} = \frac{XP}{2MA} \quad \text{and} \quad \frac{XP}{PS} = \frac{MA}{AP}$$
From this and the above observation follows
$$\begin{equation*} \frac{XS}{XP} \cdot \frac{PT}{PB} = \frac{XP}{2MA} \cdot \frac{PA}{PS} = \frac{XP}{2MA} \cdot \frac{MA}{XP} = \frac{1}{2} . \tag{1} \end{equation*}$$
Let $A'$ be the intersection point of the circle $O$ and the perpendicular bisector of the chord $BC$ such that $A$, $A'$ are on the same side of the line $BC$, and $N$ be the intersection point of the lines $A'Q$ and $CT$. Since
$$\angle NCQ = \angle TCB = \angle TCA = \angle TBA = \angle TBP$$
and
$$\angle CA'Q = \frac{\angle CAB}{2} = \frac{\angle XAP}{2} = \angle PAM = \angle SPX,$$
the triangle $NCQ$ is similar to the triangle $TBP$ and the triangle $CA'Q$ is similar to the triangle $SPX$. Therefore
$$\frac{QN}{QC} = \frac{PT}{PB} \quad \text{and} \quad \frac{QC}{QA'} = \frac{XS}{XP}$$
and hence $QA' = 2QN$ by (1). This implies that $N$ is the midpoint of the line segment $QA'$. Let the circle $O_{2}$ touch the line segment $AC$ at $Y$. Since
$$\angle ACN = \angle ACT = \angle BCT = \angle QCN$$
and $|CY| = |CQ|$, the triangles $YCN$ and $QCN$ are congruent and hence $NY \perp AC$ and $NY = NQ = NA'$. Therefore, $N$ is the center of the circle $O_{2}$, which completes the proof.