Is well-known that if we have two circumferences $\omega$ and $\Omega$ tangents internally on $X$ and there are a chord $AB$ of \Omega tangent to $\omega$ in $Y$, then $XY$ cut the arc $AB$ in its midpoint (can prove it with homothety or inversion). Let $M_a$, $M_b$, $M_c$ the midpoints of the arcs $\overarc{BAC}$, $\overarc{CBA}$, $\overarc{ACB}$. We have that $M_a$, $P_a$ and $T_a$ are collinears. By Law of sines: $\dfrac{\angle BM_aP_a}{\angle P_aM_aC}=\dfrac{BP_a\cdot M_aC}{P_aC\cdot M_aB}=\dfrac{BP_a}{P_aC}$ and we have that $\angle BM_aP_a=\angle BAT_a$, $\angle P_aM_aC=\angle T_aAC$. Then $\dfrac{\angle BAT_a}{\angle T_aAC}=\dfrac{BP_a}{P_aC}$. The similar form we have that $\dfrac{\angle ACT_c}{\angle T_cCB}=\dfrac{AP_c}{P_cB}$ and $\dfrac{\angle CBT_b}{\angle T_bBA}=\dfrac{CP_b}{P_bA}$. We conclude that by Ceva Trigonometric: $$1=\dfrac{\angle BAT_a \cdot \angle ACT_c \cdot \angle CBT_b}{\angle T_aAC \cdot \angle T_cCB \cdot \angle T_bBA}=\dfrac{BP_a \cdot AP_c \cdot CP_b}{P_aC \cdot P_cB \cdot P_bA}$$So by Ceva's Theorem we finish.