Well-Known, See China TST 2002 Quiz 6 P2

Too easy. Note that from the description $C_1$ and $C_2$ are the midpoints of arcs $AP$ and $BP$ not containing $B$ and $A$ respectively. It is known that $\omega_1$ and $\omega_2$ go through the incenter of $ABP$.

Nice and easy Since $\angle PBC_1 = \angle PC_2C_1 = \frac{\angle PC_2Q}{2} = \angle PBQ$, $C_1, Q, B$ are collinear and similarly, $B,Q,C_1$ are collinear. Then, $\angle QBA = \angle C_1BA = \angle C_1C_2A = \angle C_1C_2P = \angle C_1BP$ and so $BQ$ is angle bisector in $\triangle PAB$. Similarly, $AQ$ is also angle bisector and so $Q$ must be the incenter of $\triangle PAB$

Nice problem $PQ$ is the radical axis of $\omega _1$ and $\omega _2$. Let $M = PQ \cap C_1 C_2$, it's known that $\triangle PC_2 Q$ and $PC_1 B$ are isosceles triangles. Key Claim: $A,Q,C_2$ are collinear and $B,Q,C_1$ are collinear. Proof: $\angle PC_2 C_1 = \angle PC_2 M = \frac{\angle PC_2 Q}{2}$ $\angle PBQ = \frac{\angle PC_2 Q}{2}$ $\angle PBC_1 = \angle PC_2 C_1$ $\implies \angle PBQ = \angle PBC_1$ $\implies C_1 , Q , B$ are collinear, similarly $A,Q,C_2$ are collinear. $\angle C_1 BA = \angle C_1 C_2 A \implies \angle QBA = \angle C_1 C_2 A = \frac{\angle PC_2Q}{2} = \angle QBP$. Similarly $\angle QAB = \angle QAP$. $\implies Q$ is intersection of angle bisectors in $\triangle APB$, hence the incenter.

I usually do not check South American Contests but as Brazil was mentioned in evan sir's blog I knew that I had to see it. Note by INMO2018P3,we have $C_1,Q,B$ and $C_2,Q,A$ collinear. Since $PQ$ is the radical axes we have $\angle C_2C_1P=\angle C_2C_1Q$ and vice versa. From here the rest is just simple angle chase $ \angle QAB=\angle C_2AB =\angle C_2C_1B=\angle C_2C_1Q=\angle C_2C_1P =\angle C_2AP=\angle QAP $. Similarly $\angle QBA=\angle QBP$ $\implies Q\text{ is the incenter of } \triangle PAB$$\blacksquare$