Let $n$ be the line through $P$ which is parallel to $\ell,m$, and let $S,T$ be points on $\ell,m$ respectively such that $\angle SPT=2\angle EPF\ (1)$. Finally, let $Q$ be the fourth vertex of the parallelogram $SPTQ$. We'll prove that the point we're looking for is $Q$, and that $\angle EQF=\pi-\angle EPF\ (*)$. Since $QF\mapsto QE$ is a projective transformation, and so is the map sending a line through $Q$ to the line which makes an angle of $\pi-\angle EPF$ with it (such that the two lines have the appropriate orientation, i.e. the same as the orientation of the angle $\angle EPF$), it suffices to check $(*)$ for three particular positions of $E,F$ on $\ell,m$. We'll take the first two to be $E\to\infty,F\to\infty$, where the conclusion is easy to draw, and we'll take the third one such that $PF$ is the angle bisector of $\angle QPT$. Because of $(1),\ PE$ is the angle bisector of $\angle QPS$. Since $SE$ is the external angle bisector of $\angle PSQ$, it means that $E$ is the $P$-excenter of $PSQ$, so $QE$ must be the external angle bisector of $\angle PQS$, and thus orthogonal to the angle bisector of $\angle PQS$. This latter line is parallel to $PF$, the angle bisector of $\angle QPT$, so we have $QE\perp PF$. In the same way we show that $QF\perp PE$, so $Q$ is the orthocenter of $PEF$, and $(*)$ folows.

let circumcircle of $PEF$ intersect $\ell$ at $Q$ so we have:$\angle EQF=\alpha$ therfore all $QF$ lines are parallel to each other ,hence the locus of circumcenter will be a line,hence circumcircle pass thorugh the reflection of $P$ wrt locus of circumcenter. So We are done

amir2 wrote: let circumcircle of $PEF$ intersect $\ell$ at $Q$ so we have:$\angle EQF=\alpha$ therfore all $QF$ lines are parallel to each other ,hence the locus of circumcenter will be a line,hence circumcircle pass thorugh the reflection of $P$ wrt locus of circumcenter. So We are done You are completely wrong! Just tell me why you conclude from $QF$ moving parallel that the locus of circumcenter is a line? And of course your result is wrong for $0<\alpha<\frac{\pi}2$ because just see the two cases where one of $E,F$ goes to infinity, then the circumcenter will be the point of infinity that is perpendicular to $PF$ or $PE$ (The one that has no infinity point) But if these two points are not equal (which occurs when $0<\alpha<\frac{\pi}2$) then the line you mentioned shall be the infinity line! But of course it is not!

Take $Q$ as in grobber's post. It is enough to show that the triangle $\triangle XQE$ is similar to $\triangle YFQ$ (play with angles around $Q$ to show that $\angle EQF$ is $\pi -\alpha$). To prove the similarity notice that $\angle EXQ =\angle FYQ=\alpha$ and that the triangles $\triangle PXE$ and $\triangle FPY$ are similar. From the latter we have $XE/PY=XP/YF\Rightarrow XE/QX=QY/YF$.

I don't know if following is correct. Map $E\mapsto F$ is projective from line $l$ to line $m$. So envelope $EF$ is some conic. Since it is bounded betwen parallel lines $m$ and $l$ it is not parabola. It is quite obviously that $m$ and $l$ are tangents to this conic. Now, following is well known: Let $l$ and $m$ be tangents (they don't have to be parallel) on conic. If $F_{i}$ is focus of this conic then directed angle $\angle XF_{i}Y$ is constant, where $X$ and $Y$ are variable points on $l$ and $m$ such that $XY$ is also tangent on this conic. Useing this one of focus $F_{1}$ and $F_{2}$ is for sure not equal $P$ wich proves our statement. Note: From this we see that point $P$ need not to be betwen lines $l$ and $m$ since in this case envelope $EF$ is hyperbola. Please, give some comments. Thanks!

initialy you can prove that the locus of H(Orthocenter of triangle EPF) is a line perpendicular to lines l and m named it n.now assume that the intersecion point of circumcircle of EHF and line n is T.now prove that the distance between T and line l is equal with distance between P and line m.consequently T is a point that sees EF with angle 180-EPF.