Complex bash with $\triangle ABC$ inscribed in the unit circle, and let $Y,Z$ be the midpoints of arcs $AC,AB$ respectively, so that \begin{align*} |w|=|y|=|z|&=1 \\ a &= -\frac{yz}w \\ b &= -\frac{wz}y \\ c &= -\frac{wy}z \end{align*}Now $P$ is the intersection of line $AB$ with the line through $W$ parallel to $AC$. So $$p = \frac{ab\left(w+\frac{ac}w\right) - w\left(\frac{ac}w\right)(a+b)}{ab - w\left(\frac{ac}w\right)} = \frac{wz^2+\frac{y^2z^2}w+\frac{y^3z}w+wyz}{z^2-y^2} = \frac{z(y+z)(w^2+y^2)}{w(z+y)(z-y)} = \frac{z(w^2+y^2)}{w(z-y)}$$and similarly $$q = \frac{y(w^2+z^2)}{w(y-z)}$$Now $K$ is the intersection of the perpendicular bisector of $\overline{BW}$ with line $BC$, so $$\overline{k} = \frac{k}{bw} = \frac{b+c-k}{bc} \implies k = \frac{w(b+c)}{c+w} = \frac{w\left(-\frac{wz}y-\frac{wy}z\right)}{w-\frac{wy}z} = \frac{w(y^2+z^2)}{y(y-z)}$$Similarly, $$\ell = \frac{w(y^2+z^2)}{z(z-y)}$$Then we find the vector $$k-q = \frac{w(y^2+z^2)}{y(y-z)} - \frac{y(w^2+z^2)}{w(y-z)} = \frac{z^2(w+y)(w-y)}{wy(y-z)}$$Thus, let $T$ be the intersection of line $AW$ and line $KQ$. Since $T$ lies on line $AW$, we have $$\overline{t} = \frac{a+w-t}{aw} = \frac{tw-w^2+yz}{wyz}$$Since $T$ lies on line $KQ$, we have $$\frac{k-t}{k-q} \in \mathbb{R}$$$$\frac{w\left[w(y^2+z^2) - ty(y-z)\right]}{z^2(w+y)(w-y)} = -\frac{(y^2+z^2) + \overline{t}wz(y-z)}{(w+y)(w-y)}$$$$w^2(y^2+z^2) - twy(y-z) = -z^2(y^2+z^2) - \overline{t}wz^3(y-z)$$$$\overline{t} = \frac{twy(y-z) - (w^2+z^2)(y^2+z^2)}{wz^3(y-z)}$$Setting the two equations equal gives $$\frac{tw-w^2+yz}{wyz} = \frac{twy(y-z) - (w^2+z^2)(y^2+z^2)}{wz^3(y-z)}$$$$twz^2(y-z) - z^2(y-z)(w^2-yz) = twy^2(y-z) - y(w^2+z^2)(y^2+z^2)$$$$tw(y-z)^2(y+z) = w^2y^3+w^2z^3+y^3z^2+y^2z^3$$$$t = \frac{w^2y^2-w^2yz+w^2z^2+y^2z^2}{w(y-z)^2}$$Since this coordinate is symmetric in $y,z$, we see that line $LP$ passes through this point $T$ as well. $\blacksquare$