Let $S$ be the intersection of the lines $QR$ and $BC$.

First, note that the homothety with center $Q$ which transforms $\omega$ into $\Gamma$ maps $P$ to a point in $\Gamma$ whose tangent is parallel to $BC$, namely, $M$. Then, $Q$, $P$, $M$ are collinear. Then, the equality $BQM = MQC$ implies that $QP$ is the internal bisector of $BQC$ and, since $PQR = 90^\circ$, it follows that $QS$ is the external bisector of $BQC$.

Setting $BC = a$, $CA = b$, $AB = c$, and $2p = a+b+c$, we have that $BP = P'C = p-b$, $PC = p-c$ and $PP' = b-c$ (in case $b > c$; the other case is similar), and, by the angle bisector theorem, we have:
$$\frac{BS}{CS} = \frac{BQ}{CQ} = \frac{BP}{CP} = \frac{p-b}{p-c}$$
Then,
$$\frac{BS}{BC} = \frac{BS}{CS - BS} = \frac{p-b}{b-c}$$
or, equivalently, $BS = \frac{a(p-b)}{b-c}$.

Let $S$ be the area, $r$ the inradius, and $r_A$ the $A$-exradius of the triangle $ABC$.
We have that $RPS = PQS = P'J = 90^\circ$; hence, $PQS = RPS = P'JP$. Therefore, $PSR \sim P'JP$, which implies that:
$$\frac{SP}{RP} = \frac{P'J}{PP'} \quad \text{or, equivalently,}\quad \frac{\frac{a(p-b)}{b-c} + (p-b)}{2r} = \frac{P'J}{b-c}$$
and, as a consequence,
$$P'J = \frac{(a+b-c)(p-b)}{2r} = \frac{(p-c)(p-b)}{S/p} = \frac{p(p-b)(p-c)}{S} = \frac{S}{p-a} = r_A$$
(here, we use that $S = \sqrt{p(p-a)(p-b)(p-c)}$).

Therefore, $J$ is the center of the excircle of $ABC$ relative to the vertex $A$.

Similarly, it can be proved that $P'Q'$ passes through the incenter $I$ of $ABC$. (To do this, it suffices to show that $Q'P'$ is the internal bisector of $BQ'C$ and that, if $R'$ is the point diametrically opposite to $P'$ in $\omega'$, then $R'Q'$ is the external bisector of $BQ'C$. Then, if $S'$ is the intersection point of $R'Q'$ and $BC$ we can show as before that $P'IP \sim P'S'R'$).

Now we will prove that both lines $JP$ and $IP'$ pass through the midpoint $T$ of the altitude $AH$ of the triangle $ABC$. This finishes the problem, since $T = N$ would be the intersection of the lines $JP = PQ$ and $IP' = P'Q'$, and $AT$ is perpendicular to $BC$. The homothety with center $A$ that transforms $\omega$ into $\omega'$, transforms the diameter $RP$ into the diameter $P'R'$. Since $AHP \sim R'P'P$ (since $AH \parallel P'R'$), and $PT$, $PJ$ are medians relative to the corresponding parallel sides $AH$ and $P'R'$, then $T$, $P$ and $J$ are collinear. In addition, since $AHP' \sim RPP'$ (because $AH \parallel RP$) and $P'I$, $P'T$ are medians relative to the corresponding parallel sides $AH$ and $RP$, then $T$, $P'$, $I$ are also collinear. Therefore, $JP$ and $IP'$ both pass through $T$, as we wanted to prove.