Clearly $P \in \odot(AEF).$ Let $S$ be the 2nd intersection of $\odot(AEF)$ and $\omega,$ i.e. center of the spiral similarity that takes $\overline{EL}$ into $\overline{FK}.$ By Reim's theorem for $\odot(AEF)$ and $\omega$ with common chord $AS,$ we get that $S,P,D$ are collinear. Since $\odot(DEF)$ and $\omega$ are internally tangent at $D,$ it follows that $DE,DF$ bisect $\angle ADL$ and $\angle ADK.$ Thus, by angle bisector theorem, we deduce that $\frac{DL}{DA}=\frac{EL}{EA} \ , \ \frac{DK}{DA}=\frac{FK}{FA} \Longrightarrow \frac{DL}{DK}=\frac{EL}{FK}=\frac{SE}{SF}=\frac{ \sin \widehat{SPE}}{\sin \widehat{SPF}} \Longrightarrow$ $\frac{DL}{DK}=\frac{\sin \widehat{DAL}}{\sin \widehat{DAK}}=\frac{ \sin \widehat{SPE}}{\sin \widehat{SPF}}=\frac{DE}{2R_4} \cdot \frac{2R_3}{DF}=\frac{R_3}{R_4} \cdot \frac{2R_2 \cdot \sin \widehat{DAL}}{2R_1 \cdot \sin \widehat{DAK}} \Longrightarrow$ $R_1 \cdot R_4=R_2 \cdot R_3.$

According to the law of sines, it is sufficient to prove that: $ \sin ( \angle AFD).\sin ( \angle PED)=\sin ( \angle AED).\sin ( \angle PFD) $. We have: $\angle PED = \angle EDL $ and according to this one http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1967516&sid=2ea9f7046fedc0f5bf3328b258f7a07f#p1967516, $\angle EDL = \angle EDA $, so $ \angle PED = \angle EDA $ . And similarly, $ \angle PFD = \angle ADF $ . $%Error. "Arrow" is a bad command. $ $ \frac{\sin (\angle PED ) }{\sin ( \angle AED )}=\frac{\sin (\angle EDA ) }{\sin ( \angle AED) }=\frac{AE}{AD}=\frac{AF}{AD}=\frac{\sin (\angle ADF) }{\sin ( \angle AFD )}=\frac{\sin (\angle PFD ) }{\sin ( \angle AFD) } $. QED