Solved with Flow25 Denote $\omega_1$ and $\omega_2$ as the circumcircles $ADB$ and $ADC$ respectively. Let $X$ be the intersection of the perpendicular bisector of $AD$ with $BC$. It is known that $X$ is the center of the Apollonius' circle of the triangle $ABC$ with the ratio $AB/AC$, therefore $XB\cdot XC = XD^2 = XA^2$. By the inversion in $X$ with radius $XC$ we get that $B \rightarrow C$ and points $D$ and $A$ are fixed, which gives us $\omega_1 \rightarrow \omega_2$. So the tangents of $\omega_1$ and $\omega_2$ must intersect at $X$. By easy angle chasing we can see that the triangles $BKD$ and $DLC$ are similar, hence by homothety in $X$ the triangles $MPN$, $BKD$ and $DLC$ are similar (homothetic), which gives us the circumcircle of $MNP$ is tangent to $\ell$. Remark. we can just take points $M, N,P$ on $BD$, $DC$ and $KL$ respectively such that they divide these segments in equal ratio.