I believe there is an error in the statement, as $\angle B=90$ implies $AC$ is the hypotenuse of triangle $ABC$, so $AB>AC$ cannot happen.

My apologies, it should be $AB>BC$.

I think the problem statement is not true. I checked on geogebra also, it is not coming out to be true. Please check

g0USinsane777 wrote: I think the problem statement is not true. I checked on geogebra also, it is not coming out to be true. Please check I'm so sorry, it should be $BP=BC$ and not $BP=PC$. I have checked the problem character for character already and it should be correct now.

Let $\angle PBC=2\alpha$. Then $\angle BCP=\angle CPB=90^\circ -\alpha$ and $\angle ABP=90^\circ -2\alpha$, so $\angle PAQ=\angle PAB=2\alpha$, and therefore $\angle AQP=\angle QPA=90^\circ -\alpha =\angle BCP$. Hence $BCPQ$ is cyclic and $M$ is its circumcentre, so $\angle BMP=2\angle BCP=180^\circ -2\alpha =180^\circ -\angle PAB$, and therefore $ABMP$ is cyclic, so $AM$ is the angle bisector of $\angle PAB$. Hence if $D$ is the midpoint of $AB$ we have $\angle MDB=2\angle MAB=\angle PAB$, which means that $MD\parallel AP$, as desired.

Solution with coordinate bash: Let $B=(0,0)$, $A=(0,a)$, $C=(c,0)$. $P=(x,y)$ must satisfy $x^2+y^2=c^2$ for the length condition and $\frac{y-a}{x}=-\frac{x}{y}$ for the right angle condition. Solving this system results in $$P=(x,y)=\bigg(\frac{c\sqrt{a^2-c^2}}{a},\frac{c^2}{a}\bigg)$$ We can find that the length of $AP$ is $\sqrt{a^2-c^2}$ because of some really nice cancellations, so $Q=(0,a-\sqrt{a^2-c^2})$. Then $M=\bigg(\frac{c}{2},\frac{a-\sqrt{a^2-c^2}}{2}\bigg)$. We can find that the slope of $AP$ is $-\frac{\sqrt{a^2-c^2}}{c}$, so the line in question in the problem is given by the equation $$y-\frac{a-\sqrt{a^2-c^2}}{2}=-\frac{\sqrt{a^2-c^2}}{c}\bigg(x-\frac{c}{2}\bigg)$$ Substituting $x=0$ results in $y=\frac{a}{2}$, so we are done.