I think there’s a typo. With how it’s written now, P is the point at infinity (G on CD, CG and AB intersect at P)

I'll have a quick sketch. Notice that $ABFE$ is a rectangle, then $BF = AE$. Moreover, $BENF$ is an isosceles trapezoid. Therefore, we can prove that $ABNE$ is a kite. Let $EN \cap BC = K$. One can prove that $NKCG$ is a kite. Now, we can prove that $\triangle ABE \sim \triangle DGF$. Thus, \[ \frac{AB}{AE} = \frac{GD}{DH} \]One can prove that $FK = KC$ since $\triangle FNC$ is a right angled triangle. Thus, we have $\frac{\frac{AB}{2}}{BF} = \frac{GC}{CF} $. Thus, we must have $AB = 2BP$. It suffices to prove that $P$ is on the perpendicular bisector of $MN$, which is just a simple similarity stuff.

We will use cartesian coordinates, WLOG, assume that $AB=1, AD=a$ where $a>1$, let $D=(0,0), C=(1,0), B=(1,a), A=(0,a)$, we then have \[AC: y=-ax+a, BE: y=\frac{1}{a}x+a-\frac{1}{a}\]by intersecting these two lines we get $M=\left(\frac{1}{1+a^2}, \frac{a^3}{1+a^2}\right)$, also $E$ lying on line $BE$ has coordinate $\left(0,a-\frac{1}{a}\right)$. Then, $(ABFE)$ has its radius located at $\left(\frac{1}{2}, a-\frac{1}{2a}\right)$ and \[(ABEF): \left(x-\frac{1}{2}\right)^2+\left(y-a+\frac{1}{2a}\right)^2=\frac{1}{4}+\frac{1}{4a^2}\]intersecting it with line $y=-ax+a$ gives $N=\left(\frac{2}{1+a^2}, \frac{a^3-a}{1+a^2}\right)$. Next, the circumcircle of the triangle with vertices $D=(0,0), E=\left(0,a-\frac{1}{a}\right), N=\left(\frac{2}{1+a^2},\frac{a^3-a}{1+a^2}\right)$ has its $y$-coordinate, $\frac{a^2-1}{2a}$, let $j$ and $r$ be the $x$-coordinate and the radius. We have \[j^2+\left(\frac{a^2-1}{2a}\right)^2=r^2\]\[\left(\frac{2}{1+a^2}-j\right)^2+\left(\frac{a^3-a}{1+a^2}-\frac{a^2-1}{2a}\right)^2=r^2\]subtracting gives $\left(\frac{2}{1+a^2}-j\right)^2-j^2=\left(\frac{2}{1+a^2}\right)\left(\frac{2}{1+a^2}-2j\right)=\left(\frac{a^2-1}{2a}\right)^2-\left(\frac{(a^2-1)^2}{2a(a^2+1)}\right)^2 \implies j=\frac{3-a^2}{4}$. Therefore, \[(DEN): \left(x-\frac{3-a^2}{4}\right)^2+\left(y-\frac{a^2-1}{2a}\right)^2=\left(\frac{3-a^2}{4}\right)^2+\left(\frac{a^2-1}{2a}\right)^2\]and let $y=0$, we see that $x_{G}=\frac{3-a^2}{2}$, $G=\left(\frac{3-a^2}{2}, 0\right)$. We can then find that \[GF: y=\frac{2}{a}x+a-\frac{3}{a}\]and intersecting it with $y=a$ gives $P=\left(\frac{3}{2}, a\right)$. Now, it is suffices to check that \[PM^2=PN^2\]\begin{align*} &\left(\frac{3}{2}-\frac{1}{1+a^2}\right)^2+\left(a-\frac{a^3}{1+a^2}\right)^2=\left(\frac{3}{2}-\frac{2}{1+a^2}\right)^2+\left(a-\frac{a^3-a}{1+a^2}\right)^2 \\ \iff& \left(\frac{3}{2}-\frac{1}{1+a^2}\right)^2-\left(\frac{3}{2}-\frac{2}{1+a^2}\right)^2=\left(a-\frac{a^3-a}{1+a^2}\right)^2-\left(a-\frac{a^3}{1+a^2}\right)^2 \\ \iff& \left(3-\frac{3}{1+a^2}\right)\left(\frac{1}{1+a^2}\right)=\left(\frac{a}{1+a^2}\right)\left(2a-\frac{2a^3-a}{1+a^2}\right) \\ \iff& \frac{3a^2}{(1+a^2)^2}= \frac{3a^2}{(1+a^2)^2}. \ \blacksquare \end{align*}

First ,$NF//BE$ as $\angle FNA=90$. Let $NF$ intersect $AB$ at $R$. Let $I$ be the midpoint of $AB$ and $Q$ be the midpoint of $EF$. Let $FN$ intersect $AD$ at $W$. So, Clearly $BQ$ also intersect at $W$ becuz $EWFB$ is a $//$gm. $\frac{NB}{NG}=\frac{NA}{NC}$ and $\frac{NA}{NC}=\frac{NW}{NF}$ which means $BQ//FG$. Also , $IE//BQ//FG$,so , $\angle IEF=\angle EFG$. Also $\angle IEF=\angle IFE$ Bcuz $IQ$ is perp. bisector of $EF$. Thus, $\angle IFB=\angle GFC$ which means $\triangle IFB \sim \triangle GFC$ . So, $\frac{IB}{GC}=\frac{BF}{CF}$. Also, its known due to parallels that $\frac{PB}{GC}=\frac{BF}{CF}$. This means, $PB=IB$. Let $NF$ meets $AB$ at $K$, So , $BM$ is midline in $\triangle ANK$, So, $BR=AB$ , Thus, $P$ is the midpoint of $KB$. Now , Drop a perp. from $P$ on $AC$ at $J$ , So , $J$ is the midpoint too of $MN$ Bcuz $BMNK$ is Trapezoid. So, $PM=PN$ is proved.