Let $IR \cap PQ=S$ then note that its enough to show that the polar of $S$ w.r.t. $\omega$ is $BD$ which makes the problem purely projective so we make a projective transform that sends $ABCD$ into a square which means that $AC \cap BD=I$ and $AC \perp BD$, now $P$ is the infinity point of $AB,CD$ and $Q$ is the infinity point of $BC,AD$, $X$ becomes the intersection of $PI$ with the external angle bisector of $\angle ABD$ and $Y$ becomes the intersection of $IQ$ with the external angle bisector of $\angle BDA$, now note that $X,Y$ and $P,Q$ are pairs of symetry w.r.t. $AC$ so $A,I,C,R,S$ are colinear, now that means $S$ is sent to the infinity point of $AC$, but then that means $I,S$ are inverses w.r.t. $\omega$ and since $IS \perp BD$ we get that the polar of $S$ w.r.t. $\omega$ is indeed $BD$ thus we are done

solution without transformations (ou solução namoral) Monge D'alembert on the three circles of the statement gives that $S = XY \cap BD$ lies on $PQ$. If $T = IR \cap PQ$, since $IT$, $PY$ and $QX$ concur, we have $(Q,P;T,S) = -1 \Rightarrow T$ lies on $AC$ since $AT$, $PD$ and $QB$ must concur. Let $AB, BC, CD, DA$ touch $\omega$ on $E, F, G, H$ respectively and let $EG \cap FH = U$. Claim 1: $U = BD \cap AC$ (Newton's Theorem). Proof: Pascal on $(GGFHHE)$ and $(FFGEEH)$ gives that $U$ lies on $BD$, and changing $E,F,G,H$ by $H,E,F,G$ respectively gives that $U$ lies on $AC$. Claim 2: $EF$, $GH$, $PQ$ and $AC$ concur. Proof: See that $\triangle EGP$ and $\triangle FHQ$ are perspective to each other since $EG \cap FH = U$, $EP \cap FQ = B$, $GP \cap HQ = D$ are collinear, thus $EF$, $GH$ and $PQ$ concur by Desargues Theorem. Pascal on $(FFEGGH)$ gives $AC$ also concur in this point. Since $T$ lies on the polar of $B$ and $D$ (i.e, $EF$ and $GH$) wrt $\omega$, $BD$ is the polar of $T$ by La Hire and then $IR \perp BD$ as we wanted.