The nine point circle exists in all triangles , but for this answer , we will be focusing on the case when $\triangle ABC$ is acute .
First , let us remember the properties of cyclic quadrilaterals , as this will help us prove the concyclicity of points .
The main points to remember , are that the opposite angles are supplementary , and that the chords subtend equal angles at the circumference . The equal point to note is that the converse is true as well. This is how we will be proving that the points lie on a circle.
For the problem , it may be best to solve it in steps .
Step $1$:-
In step $1$ , we can perhaps prove that the feet of the altitudes lie on the same circle as the midpoints. 
To do this , a single altitude should suffice . As, if we can prove that this lies on the same circle as the midpoints , the others must as well. To prove this , maybe the midpoint theorem , and basic angle chasing will help...
Hint:-
Prove $\triangle FAX$ is isosceles using the converse of the midpoint theorem . Then prove that $\angle FXE$ and $\angle FDE$ are supplementary .
Step $2$:-
I assume you have finished step $1$ . Now , we use the information that the feet of the altitudes and the midpoints lie on the same circle , to prove that the midpoints of the lines joining the vertices to the orthocentre also lie on the same circle .
Again , properties of altitudes and midpoints should help us solve this . Note that in the figure , $M’$ is the midpoint of the line joining vertex $C$ to orthocentre $H$ . $M$ is the midpoint of $BC$
Hint:-
$M’M$ joins the midpoints of two line segments ! Use the midpoint theorem . Also , make a great observation. Since $\angle CFB = \angle BEC = 90 $ , $CB$ is the diameter of the circle which inscribes cyclic quadrilateral $CEFB$ ! Also , $M$ is the midpoint of the diameter... By angle chasing , and using these observations , you should be able to prove that $\angle FM’M = \angle FEM$, proving that the quadrilateral is cyclic
This completes our proof , as it must follow that all the other midpoints of the lines joining the vertices with the orthocentre are concyclic as well.
Ofcourse , we have proved the existence of the $9$-point circle for only acute angled triangles , but I trust you should now be able to prove it for all other triangles as well!
Here is a possible picture for the already solved problem:
Well, how could you fill in the details without a faithful picture?! (And what is the "reflection of $N$ across $S$" by construction, the point $N$ is reflected with respect to the center $S$ or conversely?!)
Best Answer
Since $QN\parallel PM\perp \ell$, it is enough to show that $QN=PM$.
$P$ and $Q$ lie on the sides of the medial triangle of $ABC$.
Let $M_2$ be the symmetric of $C$ wrt $M$, let $N_2$ be the symmetric of $B$ wrt $N$ and let us focus on the pentagon $AM_2 DE N_2$: we know that $M_2 D\parallel N_2 E\perp DE$ and $AN_2\perp AD, AM_2\perp AE$. Both $AEDM_2$ and $AN_2 ED$ are cyclic quadrilaterals since they have opposite right angles. This means that both $M_2$ and $N_2$ lie on the circumcircle of $ADE$. Since $\angle M_2 AD = \angle EAN_2$ it follows that $M_2 D = N_2 E$ and $NQ=MP$ as wanted.