Let $\gamma$ be the incircle of the triangle $ABC$ and let $A_1, B_1, C_1$ be its contact points with the sides $BC, CA, AB$, respectively. Let further $X_A$ be the point of contact of the circles $\gamma$ and $\omega_A$. The latter circle is the image of the former under a homothety centered at $X_A$. This homothety sends $A_1$ to a point $M_A$ on $\omega_A$ such that the tangent to $\omega_A$ at $M_A$ is parallel to $BC$. Consequently, $M_A$ is the midpoint of the arc $BC$ of $\omega_A$ not containing $X_A$. It follows that the angles $M_A X_A B$ and $M_A BC$ are congruent, so the triangles $M_A B A_1$ and $M_A X_A B$ are similar: $M_A B / M_A X_A = M_A A_1 / M_A B$. Rewrite the latter $M_A B^2 = M_A A_1 \cdot M_A X_A$ to deduce that $M_A$ lies on the radical axis $\ell_B$ of $B$ and $\gamma$. Similarly, $M_A$ lies on the radical axis $\ell_C$ of $C$ and $\gamma$.

Define the points $X_B, X_C, M_B, M_C$ and the line $\ell_A$ in a similar way and notice that the lines $\ell_A, \ell_B, \ell_C$ support the sides of the triangle $M_A M_B M_C$. The lines $\ell_A$ and $B_1 C_1$ are both perpendicular to $AI$, so they are parallel. Similarly, the lines $\ell_B$ and $\ell_C$ are parallel to $C_1 A_1$ and $A_1 B_1$, respectively. Consequently, the triangle $M_A M_B M_C$ is the image of the triangle $A_1 B_1 C_1$ under a homothety $\Theta$. Let $K$ be the center of $\Theta$ and let $k = M_A K / A_1 K = M_B K / B_1 K = M_C K / C_1 K$ be the similitude ratio. Notice that the lines $M_A A_1, M_B B_1$ and $M_C C_1$ are concurrent at $K$.

Since the points $A_1, B_1, X_A, X_B$ are concyclic, $A_1 K \cdot K X_A = B_1 K \cdot K X_B$. Multiply both sides by $k$ to get $M_A K \cdot K X_A = M_B K \cdot K X_B$ and deduce thereby that $K$ lies on the radical axis $CC'$ of $\omega_A$ and $\omega_B$. Similarly, both lines $AA'$ and $BB'$ pass through $K$.
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Finally, consider the image $O'$ of $I$ under $\Theta$. It lies on the line through $M_A$ parallel to $A_1 I$ (and hence perpendicular to $BC$); since $M_A$ is the midpoint of the arc $BC$, this line must be $M_A O$. Similarly, $O'$ lies on the line $M_B O$, so $O' = O$. Consequently, the points $I, K$ and $O$ are collinear.

Remark 1. Many steps in this solution allow different reasonings. For instance, one may see that the lines $A_1 X_A$ and $B_1 X_B$ are concurrent at point $K$ on the radical axis $CC'$ of the circles $\omega_A$ and $\omega_B$ by applying Newton's theorem to the quadrilateral $X_A X_B A_1 B_1$ (since the common tangents at $X_A$ and $X_B$ intersect on $CC'$). Then one can conclude that $K A_1 / K B_1 = K M_A / K M_B$, thus obtaining that the triangles $M_A M_B M_C$ and $A_1 B_1 C_1$ are homothetical at $K$ (and therefore $K$ is the radical center of $\omega_A, \omega_B$, and $\omega_C$). Finally, considering the inversion with the pole $K$ and the power equal to $K X_1 \cdot K M_A$ followed by the reflection at $P$ we see that the circles $\omega_A, \omega_B$, and $\omega_C$ are invariant under this transform; next, the image of $\gamma$ is the circumcircle of $M_A M_B M_C$ and it is tangent to all the circles $\omega_A, \omega_B$, and $\omega_C$, hence its center is $O$, and thus $O, I$, and $K$ are collinear.

Remark 2. Here is an outline of an alternative approach to the first part of the solution. Let $J_A$ be the excenter of the triangle $ABC$ opposite $A$. The line $J_A A_1$ meets $\gamma$ again at $Y_A$; let $Z_A$ and $N_A$ be the midpoints of the segments $A_1 Y_A$ and $J_A A_1$, respectively. Since the segment $IJ_A$ is a diameter in the circle $BCZ_A$, it follows that $BA_1 \cdot CA_1 = Z_A A_1 \cdot J_A A_1$, so $BA_1 \cdot CA_1 = N_A A_1 \cdot Y_A A_1$. Consequently, the points $B, C, N_A$ and $Y_A$ lie on some circle $\omega_A'$.

It is well known that $N_A$ lies on the perpendicular bisector of the segment $BC$, so the tangents to $\omega_A'$ and $\gamma$ at $N_A$ and $A_1$ are parallel. It follows that the tangents to these circles at $Y_A$ coincide, so $\omega_A'$ is in fact $\omega_A$, whence $X_A = Y_A$ and $M_A = N_A$. It is also well known that the midpoint $S_A$ of the segment $IJ_A$ lies both on the circumcircle $ABC$ and on the perpendicular bisector of $BC$. Since $S_A M_A$ is a midline in the triangle $A_1 I J_A$, it follows that $S_A M_A = r / 2$, where $r$ is the radius of $\gamma$ (the inradius of the triangle $ABC$). Consequently, each of the points $M_A, M_B$ and $M_C$ is at distance $R + r / 2$ from $O$ (here $R$ is the circumradius). Now proceed as above.
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