Let $M$ be the intersection of the lines $PE$ and $BC$, $N$ be the intersection of the lines $PF$ and $BC$. We will prove that $MD = ND$.

The power of $B$ with respect to the circumcircle of the triangle $AKD$ gives
$$BK \cdot BA = BD^2 = BF^2, i.e.\ BK^2 + BK \cdot KF + BK \cdot AF = (BK + KF)^2.$$
Therefore, $AF = \frac{KF \cdot BD}{BK}$. On the other hand as $FN \parallel KD$ we have $\frac{ND}{BD} = \frac{KF}{BK}$
and hence $ND = AF$. Similarly we can get $MD = AE$ and then $AE = AF$. Thus,
$MD = ND$.

Now note that $\angle DAE = \angle LDC = \angle EMD$ and hence the points $A, E, D, M$ are concyclic. Thus the circumradius of the triangle $EMD$ is $R_2$. Since $PD = 2R_4 \cdot \sin(\angle PED)$, $MD = 2R_2 \cdot \sin(\angle MED)$ and $\angle MED = \angle PED$, we get $\frac{PD}{MD} = \frac{R_1}{R_2}$. In a similar way we can get $\frac{R_3}{R_1} = \frac{PD}{ND}$ and the result follows.