Let the tangent drawn to the parabola $$y^2 = 24x$$ at the point $$(\alpha, \beta)$$ is perpendicular to the line $$2x + 2y = 5$$. Then the normal to the hyperbola $$\dfrac{x^2}{\alpha^2} - \dfrac{y^2}{\beta^2} = 1$$ at the point $$(\alpha + 4, \beta + 4)$$ does NOT pass through the point:

We need to find the point through which the normal to the hyperbola does NOT pass.

The parabola is $$y^2 = 24x$$, so $$4a = 24$$, giving $$a = 6$$.

A parametric point on the parabola is $$(at^2, 2at) = (6t^2, 12t)$$.

Differentiating $$y^2 = 24x$$ implicitly: $$2y \frac{dy}{dx} = 24$$, so $$\frac{dy}{dx} = \frac{12}{y} = \frac{12}{12t} = \frac{1}{t}$$.

The tangent must be perpendicular to $$2x + 2y = 5$$ (slope $$= -1$$), so the tangent slope $$= 1$$.

$$\frac{1}{t} = 1 \implies t = 1$$

So $$(\alpha, \beta) = (6, 12)$$.

Hyperbola: $$\frac{x^2}{36} - \frac{y^2}{144} = 1$$, with $$a^2 = 36$$ and $$b^2 = 144$$.

The point on the hyperbola is $$(\alpha + 4, \beta + 4) = (10, 16)$$.

Verification: $$\frac{100}{36} - \frac{256}{144} = \frac{400 - 256}{144} = \frac{144}{144} = 1$$ ✓

Differentiating $$\frac{x^2}{36} - \frac{y^2}{144} = 1$$ implicitly:

$$\frac{2x}{36} - \frac{2y}{144}\frac{dy}{dx} = 0$$

$$\frac{dy}{dx} = \frac{144x}{36y} = \frac{4x}{y}$$

At $$(10, 16)$$: $$\frac{dy}{dx} = \frac{40}{16} = \frac{5}{2}$$

The slope of the normal is $$m_n = -\frac{1}{dy/dx} = -\frac{2}{5}$$.

The normal at $$(10, 16)$$:

$$y - 16 = -\frac{2}{5}(x - 10)$$

$$5(y - 16) = -2(x - 10)$$

$$5y - 80 = -2x + 20$$

$$2x + 5y = 100$$

Option A: $$(25, 10)$$: $$2(25) + 5(10) = 50 + 50 = 100$$ ✓

Option B: $$(20, 12)$$: $$2(20) + 5(12) = 40 + 60 = 100$$ ✓

Option C: $$(30, 8)$$: $$2(30) + 5(8) = 60 + 40 = 100$$ ✓

Option D: $$(15, 13)$$: $$2(15) + 5(13) = 30 + 65 = 95 \neq 100$$ ✗

Therefore, the correct answer is Option D: $$(15, 13)$$.