We redefine $N'$ as the parallel from $A'$ to $B'C'$. Let $L=AI\cap A'N$, first of all we need to observe the fact that $AL\perp A'N'$, since $AI\perp B'C'$. We start by observing that, by the reciprocal of Menelaus, we need to demonstrate that : $$\frac{A'P}{AP}\cdot\frac{AN'}{B'N'}\cdot \frac{B'M}{MA'}=\frac{A'P}{AP}\cdot\frac{AN'}{B'N'}=1$$Which is equivalent to $$\frac{A'P}{AP}\stackrel{?}{=}\frac{B'N'}{AN'} \Leftrightarrow \frac{AP}{AN'}\stackrel{?}{=}\frac{A'P}{B'N'}$$. In order to demonstrate that, we need two claims Claim 1 $\triangle AB'P\sim\triangle AN'A'$. Notice that $APIB'$ is an inscribed quadrilateral since $\angle IB'P=\angle IPA=90^{\circ}$. So $\angle PB'A=\angle AIP=90^{\circ}-\angle IAP=\angle AA'N' $. Hence, by $AA$ $\triangle AB'P\sim\triangle AN'A'$. With this claim we obtain the fact that $\frac{AP}{AN'}=\frac{AB'}{AA'}$. Claim 2 $\frac{AB'}{AA'}=\frac{A'P}{B'N'}$. We proved earlier that $\angle PB'A=\angle PA'N$, hence $PB'N'A'$ is an inscribed quadrilateral. Now, $\triangle PMB'\sim\triangle A'MN$ hence $\frac{B'M}{MN}=\frac{B'P}{A'N}=\frac{AB'}{AA'}$. $(1)$ Also, $\triangle PMA'\sim\triangle B'MN$ so $\frac{A'P}{B'N}=\frac{A'M}{MN}=\frac{B'M}{MN}$.$ (2)$ From $(1)+(2)$ we conclude the claim. From Claim 1 +Claim 2 we obtain that $\frac{AP}{AN'}=\frac{A'P}{B'N'}$. As stated above, by the reciprocal of Menelaus theorem we obtain the fact that $N'=N$, done.