Let $D$ be the second intersection of $\tau$ and the bisector $\angle BCA$.Let $E,F$ be the tangency point
of $\Omega$ with $AB$ and $\tau$ respectively.
$Lemma$ 1: $DE \cdot DF=DB^2$
$Proof$ : $$\angle DFB= \frac{\angle C}{2}= \angle DBE$$.
So, $\triangle DFB$ is similar to $\triangle DBE$
$$\rightarrow DE \cdot DF=DB^2$$
$Lemma$ 2: $DP \cdot DQ=DB^2$
$Proof$ : By, POP we get, $$P_\Omega(D)=DP \cdot DQ=DE \cdot DF=DB^2$$
$Lemma$ 3: $BI$ is the angle bisector of $\angle PBQ$ where $I$ is the incentre of $\triangle ABC$
$Proof$ : From $Lemma$ $3$ we get $$DP \cdot DQ=DB^2$$
And, $P,Q,D$ collinear. So, $D$ is the centre of the Appolonian Circle of $\triangle BPQ$.
Let, the Appolonian Circle $(D,DB)$ intersect $PQ$ at $I$. So, $BI$ is the angle bisector of $\angle PBQ$.
And from $Fact$ $5$ we get $I$ is the incentre of $\triangle ABC$.
Now, $$\angle PBI=\angle QBI$$
$$\rightarrow \angle PBI + \angle ABI=\angle QBI + \angle CBI$$
$$\rightarrow \angle ABP= \angle QBC$$