In $\triangle ABC$, $AB = 8$, $BC = 7$, $CA = 6$ and side $BC$ is extended, as shown in the figure, to a point $P$ so that $\triangle PAB$ is similar to $\triangle PCA$.  The length of $PC$ is

[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair A=origin, P=(1.5,5), B=(8,0), C=P+2.5*dir(P--B);
draw(A--P--C--A--B--C);
label("$A$", A, W);
label("$B$", B, E);
label("$C$", C, NE);
label("$P$", P, NW);
label("$6$", 3*dir(A--C), SE);
label("$7$", B+3*dir(B--C), NE);
label("$8$", (4,0), S);[/asy]
$ \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$
1. Given that $\triangle PAB \sim \triangle PCA$, we can set up the proportion of corresponding sides. This similarity gives us the following ratio:
   \[
   \frac{PC}{PA} = \frac{CA}{AB} = \frac{6}{8} = \frac{3}{4}
   \]
   This implies:
   \[
   PA = \frac{4}{3} PC
   \]

2. We also know that:
   \[
   \frac{PA}{PC + 7} = \frac{CA}{AB} = \frac{3}{4}
   \]
   Substituting $PA = \frac{4}{3} PC$ into the equation, we get:
   \[
   \frac{\frac{4}{3} PC}{PC + 7} = \frac{3}{4}
   \]

3. To solve for $PC$, we cross-multiply to clear the fraction:
   \[
   4 \left(\frac{4}{3} PC\right) = 3 (PC + 7)
   \]
   Simplifying, we get:
   \[
   \frac{16}{3} PC = 3PC + 21
   \]

4. To eliminate the fraction, multiply every term by 3:
   \[
   16 PC = 9 PC + 63
   \]

5. Subtract $9 PC$ from both sides to isolate $PC$:
   \[
   7 PC = 63
   \]

6. Finally, divide both sides by 7 to solve for $PC$:
   \[
   PC = 9
   \]

Conclusion:
\[
\boxed{9}
\]