smc

Since we are given that $△PAB\sim △PCA$, we have $\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}$. Solving for $PA$ in $\frac{PC}{PA}=\frac{6}{8}=\frac{3}{4}$ gives us $PA=\frac{4PC}{3}$. We also have $\frac{PA}{PC+7}=\frac{3}{4}$. Substituting $PA$ in for our expression yields $\frac{\frac{4PC}{3}}{PC+7}=\frac{3}{4}$ Which we can further simplify to $\frac{16PC}{3}=3PC+21$ $\frac{7PC}{3}=21$ $PC=9⟹\boxed{9}$