We start with the following lemma.
Lemma 1. Points $M, N, P, Q$ are concyclic.
Point $M$ is the Miquel point of lines $A P=A B, P S=\ell, A S=A D$, and $B R=B C$, and point $N$ is the Miquel point of lines $C Q=C D, R C=B C, Q R=\ell$, and $D S=A D$. Both points $M$ and $N$ are on the circumcircle of the triangle determined by the common lines $A D, \ell$, and $B C$, which is $L R S$.
Then, since quadrilaterals $Q N R C, P M A S$, and $A B C D$ are all cyclic, using directed angles (modulo $180^\circ$ )
$$\begin{aligned} \measuredangle N M P & =\measuredangle N M S+\measuredangle S M P=\measuredangle N R S+\measuredangle S A P=\measuredangle N R Q+\measuredangle D A B=\measuredangle N R Q+\measuredangle D C B \\ & =\measuredangle N R Q+\measuredangle Q C R=\measuredangle N R Q+\measuredangle Q N R=\measuredangle N Q R=\measuredangle N Q P, \end{aligned}$$
which implies that $M N Q P$ is a cyclic quadrilateral.

Let $E$ be the Miquel point of $A B C D$ (that is, of lines $A B, B C, C D, D A$ ). It is well known that $E$ lies in the line $t$ connecting the intersections of the opposite lines of $A B C D$. Let lines $N Q$ and $t$ meet at $T$. If $T \neq E$, using directed angles, looking at the circumcircles of $L A B$ (which contains, by definition, $E$ and $M$ ), $A P S$ (which also contains $M$ ), and $M N Q P$,
$$\measuredangle T E M=\measuredangle L E M=\measuredangle L A M=\measuredangle S A M=\measuredangle S P M=\measuredangle Q P M=\measuredangle Q N M=\measuredangle T N M,$$
that is, $T$ lies in the circumcircle $\omega$ of $E M N$. If $T=E$, the same computation shows that $\measuredangle L E M=\measuredangle E N M$, which means that $t$ is tangent to $\omega$.
Now let lines $M P$ and $t$ meet at $V$. An analogous computation shows, by looking at the circumcircles of $L C D$ (which contains $E$ and $N$ ), $C Q R$, and $M N Q P$, that $V$ lies in $\omega$ as well, and that if $V=E$ then $t$ is tangent to $\omega$.
Therefore, since $\omega$ meet $t$ at $T, V$, and $E$, either $T=V$ if both $T \neq E$ and $V \neq E$ or $T=V=E$. At any rate, the intersection of lines $M P$ and $N Q$ lies in $t$.

Barycentric coordinates are a viable way to solve the problem, but even the solution we have found had some clever computations. Here is an outline of this solution.
Lemma 2. Denote by $\operatorname{pow}_{\omega} X$ the power of point $X$ with respect to circle $\omega$. Let $\Gamma_{1}$ and $\Gamma_{2}$ be circles with different centers. Considering $A B C$ as the reference triangle in barycentric coordinates, the radical axis of $\Gamma_{1}$ and $\Gamma_{2}$ is given by
$$\left(\operatorname{pow}_{\Gamma_{1}} A-\operatorname{pow}_{\Gamma_{2}} A\right) x+\left(\operatorname{pow}_{\Gamma_{1}} B-\operatorname{pow}_{\Gamma_{2}} B\right) y+\left(\operatorname{pow}_{\Gamma_{1}} C-\operatorname{pow}_{\Gamma_{2}} C\right) z=0 .$$
Proof: Let $\Gamma_{i}$ have the equation $\Gamma_{i}(x, y, z)=-a^{2} y z-b^{2} z x-c^{2} x y+(x+y+z)\left(r_{i} x+s_{i} y+t_{i} z\right)$. Then $\operatorname{pow}_{\Gamma_{i}} P=\Gamma_{i}(P)$. In particular, $\operatorname{pow}_{\Gamma_{i}} A=\Gamma_{i}(1,0,0)=r_{i}$ and, similarly, $\operatorname{pow}_{\Gamma_{i}} B=s_{i}$ and $\operatorname{pow}_{\Gamma_{i}} C=t_{i}$.
Finally, the radical axis is
$$\begin{aligned} & \operatorname{pow}_{\Gamma_{1}} P=\operatorname{pow}_{\Gamma_{2}} P \\ \Longleftrightarrow & \Gamma_{1}(x, y, z)=\Gamma_{2}(x, y, z) \\ \Longleftrightarrow & r_{1} x+s_{1} y+t_{1} z=r_{2} x+s_{2} y+t_{2} z \\ \Longleftrightarrow & \left(\operatorname{pow}_{\Gamma_{1}} A-\operatorname{pow}_{\Gamma_{2}} A\right) x+\left(\operatorname{pow}_{\Gamma_{1}} B-\operatorname{pow}_{\Gamma_{2}} B\right) y+\left(\operatorname{pow}_{\Gamma_{1}} C-\operatorname{pow}_{\Gamma_{2}} C\right) z=0 . \end{aligned}$$
We still use the Miquel point $E$ of $A B C D$. Notice that the problem is equivalent to proving that lines $M P, N Q$, and $E K$ are concurrent. The main idea is writing these three lines as radical axes. In fact, by definition of points $M, N$, and $E$ :
- $M P$ is the radical axis of the circumcircles of $P A S$ and $P B R$;
- $N Q$ is the radical axis of the circumcircles of $Q C R$ and $Q D S$;
- $E K$ is the radical axis of the circumcircles of $K B C$ and $K A D$.
Looking at these facts and the diagram, it makes sense to take triangle $K Q P$ the reference triangle. Because of that, we do not really need to draw circles nor even points $M$ and $N$, as all powers can be computed directly from points in lines $K P, K Q$, and $P Q$.

Associate $P$ with the $x$-coordinate, $Q$ with the $y$-coordinate, and $K$ with the $z$-coordinate. Applying the lemma, the equations of lines $P M, Q N$, and $E K$ are
- $M P:(K A \cdot K P-K B \cdot K P) x+(Q S \cdot Q P-Q R \cdot Q P) y=0$
- $N Q:(K C \cdot K Q-K D \cdot K Q) x+(P R \cdot P Q-P S \cdot P Q) z=0$
- $M P:(-Q C \cdot Q K+Q D \cdot Q K) y+(P B \cdot P K-P A \cdot P K) z=0$
These equations simplify to
- $M P:(A B \cdot K P) x+(P Q \cdot R S) y=0$
- $N Q:(-C D \cdot K Q) x+(P Q \cdot R S) z=0$
- $M P:(C D \cdot K Q) y+(A B \cdot K P) z=0$
Now, if $u=A B \cdot K P, v=P Q \cdot R S$, and $w=C D \cdot K Q$, it suffices to show that
$$\left|\begin{array}{ccc} u & v & 0 \\-w & 0 & v \\0 & w & u\end{array}\right|=0$$
which is a straightforward computation.