Let T and C respectively, be the transverse and conjugate axes of the hyperbola $$16x^2 - y^2 + 64x + 4y + 44 = 0$$. Then the area of the region above the parabola $$x^2 = y + 4$$, below the transverse axis T and on the right of the conjugate axis C is:

We need to find the area of a region defined by the hyperbola's axes and a parabola.

Rewrite the hyperbola in standard form

$$ 16x^2 - y^2 + 64x + 4y + 44 = 0 $$

$$ 16(x^2 + 4x) - (y^2 - 4y) + 44 = 0 $$

$$ 16(x^2 + 4x + 4) - (y^2 - 4y + 4) + 44 - 64 + 4 = 0 $$

$$ 16(x + 2)^2 - (y - 2)^2 = 16 $$

$$ \frac{(x+2)^2}{1} - \frac{(y-2)^2}{16} = 1 $$

Center: $$(-2, 2)$$, $$a = 1$$, $$b = 4$$.

Identify axes

Transverse axis T: $$y = 2$$ (horizontal line through center)

Conjugate axis C: $$x = -2$$ (vertical line through center)

Set up the region

Region: above parabola $$x^2 = y + 4$$ (i.e., $$y = x^2 - 4$$), below transverse axis $$y = 2$$, and to the right of conjugate axis $$x = -2$$.

Find intersection points

Parabola meets $$y = 2$$: $$x^2 = 6$$, so $$x = \sqrt{6}$$ (taking $$x > -2$$).

Also $$x = -\sqrt{6} \approx -2.449 < -2$$.

Parabola meets $$x = -2$$: $$y = 4 - 4 = 0$$.

So the region is bounded by $$x = -2$$ (left), $$y = 2$$ (top), and $$y = x^2 - 4$$ (bottom), from $$x = -2$$ to $$x = \sqrt{6}$$.

Compute the area

$$ A = \int_{-2}^{\sqrt{6}} [2 - (x^2 - 4)]\,dx = \int_{-2}^{\sqrt{6}} (6 - x^2)\,dx $$

$$ = \left[6x - \frac{x^3}{3}\right]_{-2}^{\sqrt{6}} $$

At $$x = \sqrt{6}$$: $$6\sqrt{6} - \frac{6\sqrt{6}}{3} = 6\sqrt{6} - 2\sqrt{6} = 4\sqrt{6}$$

At $$x = -2$$: $$-12 + \frac{8}{3} = -\frac{28}{3}$$

$$ A = 4\sqrt{6} - \left(-\frac{28}{3}\right) = 4\sqrt{6} + \frac{28}{3} $$

This matches Option 2: $$4\sqrt{6} + \frac{28}{3}$$.

The answer is $$\boxed{4\sqrt{6} + \frac{28}{3}}$$.