(1) Since $C$, $A$, $B$, $E$ are concyclic, and $D$, $A$, $B$, $F$ are concyclic, $CA = AD$, and by the theorem of power of a point, we have
$$CB \cdot CF = CA \cdot CD = DA \cdot DC = DB \cdot DE. \qquad \textcircled{1}$$
Suppose on the contrary that $l_1$ and $l_2$ do not intersect, then $CD \parallel EF$, hence $\frac{CF}{CB} = \frac{DE}{DB}$. Plugging into (1), we get $CB^2 = DE^2$, thus $CB = DB$, hence $BA \perp CD$. It follows that $CB$ and $DB$ are the diameters of $K_1$ and $K_2$ respectively, hence $K_1$ and $K_2$ have same radii, which contradicts with assumption. Thus $l_1$ and $l_2$ have a unique common point.

(2) Join $AE$, $AF$ and $PF$, we have
$$\angle CAE = \angle CBE = \angle DBF = \angle DAF.$$
Since $AP \perp CD$, $AP$ is the bisector of $\angle EAF$. Since $P$ is on the perpendicular bisector of the segment $EF$, $P$ is on the circumcircle of $\triangle AEF$. We have
$$\angle EPF = 180^\circ - \angle EAF = \angle CAE + \angle DAF = 2\angle CAE = 2\angle CBE.$$
Hence $B$ is on the circle with center $P$ and radius $PE$, denoting this circle by $\Gamma$. Let $R$ be the radius of $\Gamma$. By the

theorem of power of a point, we have
$$2CA^2 = CA \cdot CD = CB \cdot CF = CP^2 - R^2,$$
thus
$$AP^2 = CP^2 - CA^2 = (2CA^2 + R^2) - CA^2 = CA^2 + PE^2.$$
It follows that $CA$, $AP$, $PE$ form the side lengths of a right triangle.