Call $\{R_A\}=(PEF)\cap (PB_0C_0)$ and similarly for $R_B$ and $R_C$. Then $S_AT_A$ is the perpendicular bisector of $PR_A$, and thus their concurrency is equivalent to proving that $P,R_A,R_B,R_C$ are concyclic. Consider the inversion of pole $P$ fixing $\gamma$, and call $Q'$ the inverse of the point $Q$ in general. Notice that the cyclic quadrilateral is equivalent to $R_A',R_B',R_C'$ being collinear, which is further equivalent to proving that $\triangle D'E'F'$ and $\triangle A_0'B_0'C_0'$ are perspective. From Desargues Theorem, we shall prove that $D'A_0',E'B_0'$ and $F'C_0'$ are concurrent. Now call $\triangle XYZ$ the triangle formed by the tangents in $D',E',F'$ to $\gamma$. Due to Brianchon's theorem, $A,P,X$ are collinear, thus $A_0'=(PX)\cap \gamma$, implying that $XA_0'\cap YB_0'\cap ZC_0' = \{P\}$. But this immediately implies the conclusion via the following lemma: In $\triangle ABC$ let $\triangle DEF$ be the tangential triangle and $\gamma$ the incircle. Let $X,Y,Z\in\gamma$. Then $$AX\cap BY\cap CZ\neq \emptyset \iff DX\cap EY\cap FZ\neq \emptyset.$$(The proof can be done via a projective map or trig ceva.)