Let P be a moving point on the circle x^2 + y^2 - 6x - 8y + 21 = 0...

The equation of the given circle is $x^2 + y^2 - 6x - 8y + 21 = 0$.

The center of the circle is $C(3, 4)$ and its radius is $r = \sqrt{3^2 + 4^2 - 21} = \sqrt{4} = 2$.

The equation of the given parabola is $x^2 + 6x + y + 13 = 0$.

Rewriting the equation by completing the square, we get $(x + 3)^2 = -(y + 4)$.

The vertex of the parabola is $V(-3, -4)$.

The maximum distance of a moving point $P$ on the circle from the vertex $V$ is given by $CV + r$.

The distance between the center $C(3, 4)$ and the vertex $V(-3, -4)$ is $CV = \sqrt{(3 - (-3))^2 + (4 - (-4))^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$.

Therefore, the maximum distance is $10 + 2 = 12$.

Answer: $12$