Let p be the number of all triangles that can be formed by joining the vertices of a regular polygon P of n sides and q be the number of all quadrilaterals that can be formed by joining the vertices of P. If $$p + q = 126$$, then the eccentricity of the ellipse $$\frac{x^2}{16} + \frac{y^2}{n} = 1$$ is :

The number of triangles that can be formed from the vertices of a regular $$n$$-gon is the number of ways of choosing any $$3$$ vertices:
$$p = {}^{n}C_{3} = \frac{n(n-1)(n-2)}{6}$$

The number of quadrilaterals is the number of ways of choosing any $$4$$ vertices:
$$q = {}^{n}C_{4} = \frac{n(n-1)(n-2)(n-3)}{24}$$

We are told that
$$p + q = 126$$

Substituting the expressions for $$p$$ and $$q$$:
$$\frac{n(n-1)(n-2)}{6} + \frac{n(n-1)(n-2)(n-3)}{24} = 126$$

Multiply every term by $$24$$ to clear denominators:
$$4n(n-1)(n-2) + n(n-1)(n-2)(n-3) = 3024$$

Factor out $$n(n-1)(n-2)$$:
$$n(n-1)(n-2)\,\bigl[\,4 + (n-3)\bigr] = 3024$$

Simplify the bracket:
$$n(n-1)(n-2)(n+1) = 3024$$

Try successive integer values of $$n \ge 4$$ until the product equals $$3024$$.
Case 6: $$6\cdot5\cdot4\cdot7 = 840 \lt 3024$$
Case 7: $$7\cdot6\cdot5\cdot8 = 1680 \lt 3024$$
Case 8: $$8\cdot7\cdot6\cdot9 = 3024$$ — match found.

Hence $$n = 8$$.

The ellipse given is
$$\frac{x^2}{16} + \frac{y^2}{n} = 1 \quad\Longrightarrow\quad \frac{x^2}{16} + \frac{y^2}{8} = 1$$ because $$n = 8$$.

For an ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $$a^2 \gt b^2$$, the eccentricity is
$$e = \sqrt{1 - \frac{b^{2}}{a^{2}}}$$

Here $$a^2 = 16,\; b^2 = 8$$, so
$$e = \sqrt{1 - \frac{8}{16}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$

Therefore the required eccentricity is $$\boxed{\dfrac{1}{\sqrt{2}}}$$, which corresponds to Option D.