If the line $$y = 4 + kx, k > 0$$, is the tangent to the parabola $$y = x - x^2$$ at the point $$P$$ and $$V$$ is the vertex of the parabola, then the slope of the line through $$P$$ and $$V$$ is

We begin by finding the point of tangency. The parabola is $$y = x - x^2$$. At a point $$P = (a, a - a^2)$$, the slope of the tangent is:

$$\frac{dy}{dx} = 1 - 2a$$

Since the tangent line is $$y = 4 + kx$$ with slope $$k$$, it follows that $$k = 1 - 2a$$.

Requiring that the point $$P$$ lies on the tangent line gives $$a - a^2 = 4 + ka = 4 + (1 - 2a)a = 4 + a - 2a^2$$, which simplifies to $$a - a^2 = 4 + a - 2a^2$$.

This leads to $$a^2 = 4$$ and hence $$a = \pm 2$$.

Since $$k > 0$$, we have $$1 - 2a > 0 \Rightarrow a < \frac{1}{2}$$, so the correct choice is $$a = -2$$.

$$k = 1 - 2(-2) = 5$$ and $$P = (-2, -2 - 4) = (-2, -6)$$.

To find the vertex of the parabola, rewrite $$y = x - x^2 = -\left(x^2 - x\right) = -\left(x - \frac{1}{2}\right)^2 + \frac{1}{4}$$.

Thus, the vertex is $$V = \left(\frac{1}{2}, \frac{1}{4}\right)$$.

Finally, the slope of line $$PV$$ is $$\text{Slope} = \frac{\frac{1}{4} - (-6)}{\frac{1}{2} - (-2)} = \frac{\frac{1}{4} + 6}{\frac{1}{2} + 2} = \frac{\frac{25}{4}}{\frac{5}{2}} = \frac{25}{4} \times \frac{2}{5} = \frac{5}{2}$$.

Therefore, the slope of the line through $$P$$ and $$V$$ is $$\frac{5}{2}$$, which is Option C.