The tangents at the points $$A(1, 3)$$ and $$B(1, -1)$$ on the parabola $$y^2 - 2x - 2y = 1$$ meet at the point $$P$$. Then the area (in $$\text{unit}^2$$) of the triangle $$PAB$$ is:

We need to find the area of triangle $$PAB$$ where $$P$$ is the intersection of the tangents at $$A(1,3)$$ and $$B(1,-1)$$ on the parabola $$y^2 - 2x - 2y = 1$$. By completing the square in $$y$$, the equation becomes

$$y^2 - 2y = 2x + 1$$

which simplifies to

$$ (y-1)^2 = 2(x+1) $$

showing that the parabola has its vertex at $$(-1,1)$$ and its axis parallel to the $$x$$-axis.

To find the tangent at $$A(1,3)$$, we differentiate the original equation $$y^2 - 2x - 2y = 1$$ implicitly with respect to $$x$$. This yields

$$2y \frac{dy}{dx} - 2 - 2\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{2}{2y - 2} = \frac{1}{y-1}.$$

At the point $$A(1,3)$$, the slope of the tangent is $$\frac{1}{3-1} = \frac{1}{2}$$, so its equation can be written as $$y - 3 = \frac{1}{2}(x - 1)$$, which rearranges to $$x - 2y + 5 = 0$$.

For the tangent at $$B(1,-1)$$, substituting $$y = -1$$ into $$\frac{dy}{dx} = \frac{1}{y-1}$$ gives a slope of $$\frac{1}{-1-1} = -\frac{1}{2}$$. Thus the tangent line at $$B$$ is $$y + 1 = -\frac{1}{2}(x - 1)$$, or equivalently $$x + 2y + 1 = 0$$.

The intersection point $$P$$ of these two tangents satisfies both $$x - 2y + 5 = 0$$ and $$x + 2y + 1 = 0$$. Adding these equations gives

$$ (x - 2y + 5) + (x + 2y + 1) = 0 \implies 2x + 6 = 0 \implies x = -3. $$

Substituting $$x = -3$$ into $$x - 2y + 5 = 0$$ leads to $$-3 - 2y + 5 = 0$$, so $$y = 1$$ and hence $$P = (-3,1)$$.

Finally, the area of triangle $$PAB$$ with vertices $$P(-3,1)$$, $$A(1,3)$$, and $$B(1,-1)$$ can be computed using the determinant formula

$$\text{Area} = \frac{1}{2}\bigl|(-3)(3-(-1)) + 1((-1)-1) + 1(1-3)\bigr|. $$

$$= \frac{1}{2}\bigl|(-3)(4) + 1(-2) + 1(-2)\bigr|. $$

$$= \frac{1}{2}\bigl|-12 - 2 - 2\bigr| = \frac{1}{2} \times 16 = 8. $$

The correct answer is Option D: $$8$$.