It is known that $A,Q$ and $M$ are collinear. By Miquel's theorem on triangle $ABM$ we conclude $AD'QUF$ is cyclic, and thus: $$\angle UDI=\angle IFU=90^{\circ}-\angle UFA=90^{\circ}-\angle UQM=\angle UMQ$$$$\implies U,I,M\text{ are collinear.}$$See that $UQ\perp UM$ and $FI=ID$. Since $UI$ is the external bisector of the angle $\angle FUD$, we conclude that $UQ$ is the bisector of the same angle and therefore $B,U$ and $Q$ are collinear. By power of point from $B$ with respect to $(AUQ)$ and $(UQM)$ we conclude that $AB=BM$. Let $X$ be the midpoint of the segment $AM$. Since $ABM$ is isosceles, we have $B,I$ and $X$ collinear and $BX\perp AM$. From $\angle QUI=90^{\circ}$ we have $QUIX$ cyclic, and therefore: $$\angle QUX=\angle QIX=\angle BID=\angle FUB$$$$\implies\angle QUX=\angle FUB\implies F,U,X\text{ are collinear.}$$It's easy to see that $P,X$ and $N$ are collinear and we are done. $\blacksquare$

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