What a beautiful problem!! Let $\Gamma,\omega,I$ be the circumcircle, incircle, incenter of $P.$ Suppose that $\omega$ touches $A_iA_{i+1}$ at $T_i$ and define $M_i$ be the midpoint of $T_{i-1}T_i.$ By inversion of $\omega,$ we get $\omega_i$ gets sent to the $\omega'_i,$ which is the nine-point circle of $\triangle T_{i-1}T_iT_{i+1}$ so it's enough to prove that there exists circle tangent to $\omega'_i$ for all $i=1,2,\cdots,n.$ Let the reflection of $I$ wrt $M_iM_{i+1}$ be $I_i.$ (Here, set $X_{i+n}=X_i$ for all $i.$) claim. $I_1I_2\cdots I_n$ is cyclic. proof. Define $F_i$ be the foot of perpendicular from $I$ to $M_iM_{i+1},$ then $F_i$ is midpoint of $II_i$ hence the claim is equivalent to $F_1F_2\cdots F_n$ is cyclic. Let $M_{i-1}M_i\cap M_{i+1}M_{i+2}=L_i$ and $(A_{i-1}A_iI)\cap(A_{i+1}A_{i+2}I)=L'_i(\neq I).$ By inversion of $\omega, L_i$ gets sent to $L'_i$ so $I,L'_i, L_i$ are collinear. Also $B_i$ is the radical center of $\Gamma, (A_{i-1}A_iI)$ and $(A_{i+1}A_{i+2}I)$ thus $I, L'_i,L_i,B_i$ are collinear, in particular $IL_i\perp M_iM_{i+1}.$ Note that $(IF_{i-1}L_iF_{i+1})$ and $(IF_{i-1}M_iF_i)$ are cyclic, by simple angle chasing we have $$\angle F_iF_{i-2}F_{i-1}=90^\circ-\displaystyle\frac{\angle A_{i-2}A_{i-1}A_i+\angle A_iA_{i+1}A_{i+2}-\angle A_{i-1}A_iA_{i+1}}{2}=\angle F_iF_{i+1}F_{i-1}$$implying $F_{i-2}F_{i-1}F_iF_{i+1}$ is cyclic for all $i$ and we conclude $F_1F_2\cdots F_n$ is cyclic inductively therefore we are done. Let $O$ be the circumcenter of $I_1I_2\cdots I_n$ and $O_i$ be the center of $\omega'_i.$ Since $(IM_iT_iM_{i+1})$ and $\omega'_i$ are symmetry wrt $M_iM_{i+1},$ we get $I_i$ lies on $\omega'_i.$ Furthermore, $M_iI_{i-1}=M_iI=M_iI_i$ so $M_i$ lies on perpendicular bisector of $I_{i-1}I_i.$ Thus by angle chasing it follows that $$\angle M_iI_iO=90^\circ-\displaystyle\frac{\angle A_{i-1}A_iA_{i+1}}{2}=\angle M_iI_iO_i,$$means that $O,O_i,I_i$ are collinear. From the above, $(I_1I_2\cdots I_n)$ tangent to $\omega'_i$ and we have the desired result.

Let $I$, $O$, $\omega$, and $\Gamma$ denote the incenter, circumcenter, incircle, and circumcircle of $P$, respectively. We will prove that there exists a circle $\Omega$ with center $O$ that satisfies the condition described in the statement. Let $O_i$ be the center of $\omega_i$. Note that $\Omega$ is tangent to $\omega_i$ if and only if there exists a point on $OO_i$ in both circles. However, $OO_i$ is the perpendicular bisector of $A_iA_{i+1}$, so it intersects $\omega_i$ at the midpoints of the two arcs $A_iA_{i+1}$ of $\omega_i$. We will show that there exists $\Omega$ passing through $T_i =$ the midpoint of the arc $A_iA_{i+1}$ that does not contain $B_i$. To demonstrate this, we will show that $OT_i$ can be expressed solely in terms of $R =$ the radius of $\Gamma$, $r =$ the radius of $\omega$, and $d = OI$ (note that these three parameters are sufficient to define the relative positions of $I$, $O$, $\omega$, and $\Gamma$, which are symmetric with respect to all vertices of $P$ that we have). Recognizing that $I$ is the excenter of $\triangle A_iA_{i+1}B_i$, it is a known fact that $T_i$ is the circumcenter of $\triangle IA_iA_{i+1}$. We will establish some points that will help in the calculations. Let $M_i$ be the midpoint of the segment $A_iA_{i+1}$, $C_i$ be the point of tangency of $\omega$ on $A_iA_{i+1}$, and $D_i$ be the foot of the perpendicular from $I$ on the line $OM_i$. We will use oriented segments. (a) $R^2 = A_iM_i^2 + OM_i^2$ (b) $T_iA_i^2 = T_iI^2$ (c) $T_iA_i^2 = T_iM_i^2 + A_iM_i^2$ (d) $d^2 = OD_i^2 + ID_i^2$ (e) $T_iI^2 = T_iD_i^2 + ID_i^2$ By adding (a) + (b) - (c) - (d) + (e), we obtain $R^2 - d^2 = OM_i^2 - T_iM_i^2 - OD_i^2 + T_iD_i^2 = (OM_i - T_iM_i) \cdot (OM_i + T_iM_i) - (OD_i - T_iD_i) \cdot (OD_i + T_iD_i) = OT_i \cdot (OM_i + T_iM_i - OD_i - T_iD_i) = OT_i \cdot (2D_iM_i) = 2OT_i \cdot IC_i$ $\Rightarrow |OT_i| = \frac{|R^2 - d^2|}{2r}$

We claim there exists such a circle centered at $O,$ the circumcenter of $A_1A_2\dots A_n,$ externally tangent to all $\omega_i.$ Then, we only need to show that there exists a circle centered at $O$ externally tangent to both $\omega_1,\omega_2$ as by symmetry it will follow for all $\omega_i.$ Let $O_1,O_2$ be the centers of $\omega_1,\omega_2.$ We will have proven our claim if we show that the distance from $O$ to $OO_1\cap \omega_1$ is the same as the distance from $O$ to $OO_2\cap \omega_2,$ both times taking the intersections closest to $O.$ Call these points $M_1,M_2;$ we see that $OO_1,OO_2$ are the perpendicular bisectors of $A_1A_2,A_2A_3$ respectively, thus $M_1,M_2$ are arc midpoints of $A_1A_2,A_2A_3$ on $\omega_1,\omega_2$ not containing $B_1,B_2$ respectively. Now by Fact 5, there is a circle centered at $M_1$ passing through $A_1,A_2,$ and the incenter $I$ of $A_1A_2\dots A_n.$ Similarly, there is a circle centered at $M_2$ passing through $A_2,A_3,I.$ Thus $A_2I$ is the radical axis of these two circles, and thus $A_2I\perp M_1M_2.$ Now since $M_1O\perp A_1A_2$ and $M_2O\perp A_2A_3,$ by a simple angle chase we see that the bisectors of $\angle A_1A_2A_3$ and $\angle M_1OM_2$ are parallel. The first one is $A_2I,$ thus $M_1M_2$ is perpendicular to the angle bisector of $\angle M_1OM_2,$ so $M_1OM_2$ is isosceles and $M_1O=M_2O,$ which is what we wanted. Remark: There also exists a circle internally tangent to all $\omega_i,$ to show its existence invert about the incircle. Our circle centered at $O$ inverts to some circle $\Omega$ now tangent to all images of $\omega_i,$ and all images of $\omega_i$ have radius equal to half the inradius, as they become nine-point circles. Thus either increasing or decreasing the radius of $\Omega$ by the inradius while keeping its center the same results in a circle tangent to all images of $\omega_i$ the opposite way, and inverting back gives the result.

We claim the desired circle's center is the circumcenter. Focusing "locally" on two adjacent sides, our problem boils down to: Local Formulation wrote: Complete tangential quadrilateral $ABCD$ with $X = AB \cap CD$ and $Y = BC \cap DA$. Let $P$ be the intersection of the perpendicular bisectors of $BC$ and $CD$. Show the shortest distances from $P$ to $(BCX)$ and $(CDY)$ are equal. Denote $I$ and $J$ as the incenters of $\triangle BCX$ and $\triangle CDY$, $M$ and $N$ as the midpoints of the arcs opposite $X$ and $Y$, and $O$ as the incenter of $ABCD$. The key is to focus on $O$ as an excenter. If $\ell$ is the angle bisector of $\angle MPN$, notice \[\overline{MN} \parallel \overline{ICJ} \perp \overline{CO} \parallel \ell \implies PM = PN. \quad \blacksquare\]

shendrew7 wrote: We claim the desired circle's center is the circumcenter. Focusing "locally" on two adjacent sides, our problem boils down to: Local Formulation wrote: Complete tangential quadrilateral $ABCD$ with $X = AB \cap CD$ and $Y = BC \cap DA$. Let $P$ be the intersection of the perpendicular bisectors of $BC$ and $CD$. Show the shortest distances from $P$ to $(BCX)$ and $(CDY)$ are equal. Denote $I$ and $J$ as the incenters of $\triangle BCX$ and $\triangle CDY$, $M$ and $N$ as the midpoints of the arcs opposite $X$ and $Y$, and $O$ as the incenter of $ABCD$. The key is to focus on $O$ as an excenter. If $\ell$ is the angle bisector of $\angle MPN$, notice \[\overline{MN} \parallel \overline{ICJ} \perp \overline{CO} \parallel \ell \implies PM = PN. \quad \blacksquare\] very Nice and elegant solution for this hard problem

So people were doing brazil problems back over 200 years ago. See this here. Theorem 2 in Part 2 finishes after a homothety + Fact 5. Theorem 5 finishes, and another way is using an earlier result which says the centroid of three adjacent touch points lies on a fixed point.