First we'll prove the converse; if $K$ is the incenter of $\triangle{ABC}$ then $BL \perp PQ$. This is true because in this case $\triangle{BPK} \sim \triangle{BKQ} \sim$ intouch triangle, so $$\tfrac{\sin \angle{ABL}}{\sin \angle{CBL}} = \tfrac{tan \angle{C}/2}{\tan \angle{A}/2} \div \tfrac{\sin \angle{C}}{\sin \angle{A}} = \left(\tfrac{\cos \angle{A}/2}{\cos \angle{C}/2}\right)^2 = \tfrac{BQ}{BP} = \tfrac{\sin \angle{BPQ}}{\sin \angle{BQP}},$$and $\angle{ABC} = 180^{\circ} - \angle{PBQ}$, so $\angle{ABL} = \angle{BPQ}$ and $\angle{CBL} = \angle{BQP}$. Now back to the original problem. In order for $BL \perp PQ$, we need $\tfrac{BQ}{BP} = \tfrac{AL}{CL} \div \tfrac{AB}{BC}$ by the same method as earlier. Since $BP, AB, BC$ are constants, all we care about is the value of the expression $\tfrac{AL}{CL \cdot BQ}$. Note that as $L$ moves toward $A$, $\tfrac{AL}{CL}$ decreases and $BQ$ increases, so the whole fraction decreases, and as $L$ moves toward $C$, $\tfrac{AL}{CL}$ increases and $BQ$ decreases, so the whole fraction increases. So, each value of the fraction is only attained by one value of $L$, and since we know the incenter works, it must be the only possible value for $K$. $\square$