Solution:
Let $R$ be the midpoint of the side $AD$. Then the lines $BR$ and $PD$ are parallel. Since $\angle MAN = \angle QAP = \angle QBR = \angle QKM$, the points $A, N, K, M$ are concyclic.

Let $\ell_{4}$ be the line passing through the midpoint $W$ of the line segment $MK$ and perpendicular to the line $AN$. Let $E_{1}$ be the point of intersection of $\ell_{1}$ and $\ell_{4}$, and $E_{2}$ be the point of intersection of $\ell_{2}$ and $\ell_{3}$. We will show that the points $E_{1}$ and $E_{2}$ coincide.

Let $O$ be the circumcenter of the cyclic quadrilateral $ANKM$. $OW$ is perpendicular to the side $MK$ and $OX$ is perpendicular to the side $AN$. Hence $OW$ is parallel to $\ell_{1}$, $OX$ is parallel to $\ell_{4}$, and $XOWE_{1}$ is a parallelogram. Therefore the midpoints of the line segments $OE_{1}$ and $WX$ coincide. Similarly, the midpoints of the line segments $OE_{2}$ and $YZ$ coincide.

On the other hand, as $X, Y, Z, W$ are midpoints of the sides of the quadrilateral $ANKM$, $XYWZ$ is a parallelogram and therefore the midpoints of the line segments $WX$ and $YZ$ coincide. Hence the midpoints of the line segments $OE_{1}$ and $OE_{2}$ coincide. In other words, $E_{1}$ and $E_{2}$ are the same point, and the lines $\ell_{1}, \ell_{2}, \ell_{3}$ are concurrent.