Sol:- Note that $IN$ is symmedian in $AIC$.Let $IN \cap A_1C_1=P'$;$AI \cap A_1C_1=E$;$CI \cap A_1C_1=D$.By Iran Lemma $ADEC$ is cyclic with center $M$.Let the tangent at $I$ to $AIC$ be $k$.Note that $k$ is parallel to $A_1C_1$.$(k,IP';IE,ID)=-1$.So $P'$ is the midpoint of $ED$.Hence $MP'$ is perpendicular to $A_1C_1$.So $P' \cong P$.

GeoKing wrote: Sol:- Note that $IN$ is symmedian in $AIC$.Let $IN \cap A_1C_1=P'$;$AI \cap A_1C_1=E$;$CI \cap A_1C_1=D$.By Iran Lemma $ADEC$ is cyclic with center $M$.Let the tangent at $I$ to $AIC$ be $k$.Note that $k$ is parallel to $A_1C_1$.$(k,IP';IE,ID)=-1$.So $P'$ is the midpoint of $ED$.Hence $MP'$ is perpendicular to $A_1C_1$.So $P' \cong P$. Nice solution!

This shouldn't be too bad with Barycentrics. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then $M=(1:0:1),I=(a:b:c),C_1=(s-b:s-a:0),A_1=(0:s-c:s-b),N=(-a^2+ac:b^2:-c^2+ac)$. ($N$ is the midpoint of two excenters.) Then let $P'=NI\cap A_1C_1$ and check that $P'M\perp A_1C_1$ with strong EFFT which establishes $P'=P$, finishing. We find the coordinates of $P'$ by intersecting $$NI:\begin{vmatrix}x&y&z\\ a&b&c\\ -a^2+ac&b^2&-c^2+ac\end{vmatrix}=0$$and $$A_1C_1:\begin{vmatrix}x&y&z\\ s-b&s-a&0\\ 0&s-c&s-b\end{vmatrix}=0$$which is equivalent to solving the system \begin{align*}x+y+z&=1\\ x(-a+b+c)-y(a-b+c)+z(a+b-c)&=0\\ x(-b^2c-bc^2+abc)-y(2a^2c-2ac^2)+z(ab^2+a^2b-abc)&=0.\end{align*} Also, $\overrightarrow{A_1C_1}=(a:c-a:c)$ after some manipulation which helps a lot. If anyone wants to work out the details, please do so and post!

Let $W$ be the antipode of $N$ and $L$ the midpoint of $C_1A_1$. Now define $P'$ as $NI\cap A_1C_1$. We wish to show $MP'\parallel IW$. But this follows easily as $\triangle CNA \sim \triangle A_1BC_1$ implies $$\frac{WM}{NM}=\frac{IL}{LB}=\frac{IP'}{P'N},$$which implies our desired result.

This is equivalent to USAJMO 2014 Problem 6: https://artofproblemsolving.com/community/c5h587674p3478583 Also, I approve of the name "Duck Point" for this intersection point.

Let $I_A$ and $I_C$ be the $A$-excenter and the $C$-excenter, respectively. We know that $N$ is the midpoint of $I_CI_A$ and $I_CI_A\parallel C_1A_1$. Now, if $D=IC\cap C_1A_1$ and $E=IA\cap C_1A_1$, as it has been noted in previous solutions, $P$ is the midpoint of $DE$ by Iran's lemma. Finally, as $I-E-I_A$ and $I-D-I_C$ are collinear, we are done by homothety in $I$. $\square$

Point also appears here https://artofproblemsolving.com/community/u998255h3355708p31160241 I approve of the name "Duck Point" for this intersection point too

Consturct AI, CI cap the intouch chord and note that M is midpoint of these two intersection points and (these two points, A, C) are cyclic, then win by homothety