First, it's easy to see that $AL$ passes through the incentre $I$, which must be the midpoint of segment $PQ$. So construct the incentre $I$ by taking the midpoint of $PQ$ (through, say, perpendicular bisectors) together with the incircle $\Gamma$. Next, it's well-known that $W$ is the midpoint of arc $BC$ not containing $A$, and additionally that $WB=WI=WC$ through the incentre-excentre lemma. So construct a circle $\omega$ with centre $W$ and radius $WI$, which must pass through $B$ and $C$. Finally, it's easy to see that $B$ and $C$ must also pass through the tangents from $A$ to $\Gamma$. So construct the tangents from $A$ to $\Gamma$ by taking the midpoint $M$ of $AI$, constructing a circle $c$ with centre $M$ and radius $MA=MI$, and taking the intersections $E$ and $F$ of $c$ and $\Gamma$. As $\angle AEI = \angle AFI = 90^{\circ}$, we see that they are tangents. Then extend line $A$ past $F$ and $E$, and take the first intersections of these lines with circle $\omega$. These are points $B$ and $C$, in fact; and we are done.