I'm not sure if this is correct but we'll use the moving point lemma. Fix $A,B,E,D$ on $\Omega$ and change $C$ on $\Omega$ . the map $C \mapsto F$ is projective. Let $Y=BE \cap CF$. Clearly , $BF||CE$ so $YB=YF$. Now consider a circle with diameter $(BO)$ , So we have $OY,BF$ are concurrent on that circle. So the map $O \mapsto M$ ,where $M$ is the midpoint of $BF$ is projective which implies that the map $F \mapsto Y$ is projective. As $Y$ moves along $BE$ , $DY \cap \Omega$ is also projective. Now , clearly $BC \cap AD=P$ moves projectively. And note that $X$ is the reflection of $A$ in $PO$ which give us that the midpoint of $AX$ , lies on $PO$ and also on the circle with diameter $(AO)$. do the same as before and $C \mapsto P \mapsto X$ would be projective. So it's enough to check 3 cases for $C$ All the cases $C=B$,$C=D$,$C=E$ are easy to solve and we're done.

[Incomplete] Let $I=DE\cap BX$, $J=EF\cap BC$, $K=DF\cap CX$. By the Desargues' Theorem on $\triangle DEF$ and $\triangle XBC$, it is sufficient to prove that $I$, $J$, $K$ are collinear. From the parallel condition we get $CE\parallel BF$. By pascal on $FBXCED$ we get $\infty, K, I$ are collinear. But still remain to show $JK\equiv I\infty$.

Tafi_ak wrote: Let $I=DE\cap BX$, $J=EF\cap BC$, $K=DF\cap CX$. By the Desargues' Theorem on $\triangle DEF$ and $\triangle XBC$, it is sufficient to prove that $I$, $J$, $K$ are collinear. From the parallel condition we get $CE\parallel BF$. By pascal on $BFDECX$ and $FBXCED$ we get $\infty, K, J$ and $\infty, K, I$ are collinear $\implies I, J, K$ are collinear. I think you've made a mistake. You used Pascal twice on the same hexagon. Besides, you didn't use the condition $PA=PX$.

F_Xavier1203 wrote: I think you've made a mistake. You used Pascal twice on the same hexagon. Besides, you didn't use the condition $PA=PX$. I don't know how I made that kind of stupid Big mistake. I think I thought the some hexagon (by mistake I wrote $BFDECX$) works to show $\infty, K, J$ are collinear but unfortunately it doesn't work.

Easiest Problem I have ever seen Let $(EOB)$ intersect $BC$ at $H$. So, It is obvious that $BE=FC$. $\angle OHC=\angle OEB=\angle OCF$. So, $(COHF)$ is cyclic. And, $\angle OHC=\angle OHP=\angle ODA$, As, $\hat AD=\hat BE=\hat CF$. Let $BE$ and $CF$ intersect at $G$. So, as the inverse of $G$ is $H$ and $H$ lies on $(OPD)$ and $(OPDX)$ is cyclic by obvious angle chase. So, $G$ lies on the image of $(OPDX)$ wrt to $(ABCD)$. So, $BE$,$CF$ and $DX$ are concurrent.

Let $O,Q,R$ be the center of $\Omega,AC\cap BD,AB\cap CD,$. By DDIT($A,D,Q,R$ and $\mathcal{L}_\infty$), the infinity points of $\{AB,BD\},\{AC,CD\},\{AD,QR\}$ are involution. Note that $AX\perp OP\perp EF$. Projection by A on $\Omega$: $\{B,E\},\{C,F\},\{D,X\}$ are involution. Q.E.D.