If $$a = \displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} \dfrac{2n}{n^2 + k^2}$$ and $$f(x) = \sqrt{\dfrac{1-\cos x}{1+\cos x}}$$, $$x \in (0, 1)$$, then:

We need to find the value of $$a$$ first, then evaluate $$f$$ and $$f'$$ at $$\dfrac{a}{2}$$.

First, to find $$a$$, we evaluate the limit $$a = \lim_{n \to \infty} \sum_{k=1}^{n} \dfrac{2n}{n^2 + k^2} = \lim_{n \to \infty} \sum_{k=1}^{n} \dfrac{2}{n} \cdot \dfrac{1}{1 + \left(\frac{k}{n}\right)^2}$$

This is a Riemann sum for $$\int_0^1 \dfrac{2}{1+t^2}\,dt$$.

From this, $$a = 2\Big[\tan^{-1}(t)\Big]_0^1 = 2 \cdot \dfrac{\pi}{4} = \dfrac{\pi}{2}$$

Next, simplifying $$f(x)$$, we have $$f(x) = \sqrt{\dfrac{1 - \cos x}{1 + \cos x}}$$

Using the identities $$1 - \cos x = 2\sin^2\dfrac{x}{2}$$ and $$1 + \cos x = 2\cos^2\dfrac{x}{2}$$:

$$f(x) = \sqrt{\dfrac{2\sin^2(x/2)}{2\cos^2(x/2)}} = \left|\tan\dfrac{x}{2}\right| = \tan\dfrac{x}{2}$$

(since $$x \in (0, 1) \subset (0, \pi)$$, so $$\tan(x/2) > 0$$).

Then $$f\left(\dfrac{\pi}{4}\right) = \tan\dfrac{\pi}{8}$$

Since $$f(x) = \tan\dfrac{x}{2}$$, we get $$f'(x) = \dfrac{1}{2}\sec^2\dfrac{x}{2}$$

Thus $$f'\left(\dfrac{\pi}{4}\right) = \dfrac{1}{2}\sec^2\dfrac{\pi}{8} = \dfrac{1}{2\cos^2(\pi/8)}$$

Next, we verify the relationship by checking Option C: $$\sqrt{2}\,f\left(\dfrac{a}{2}\right) = f'\left(\dfrac{a}{2}\right)$$.

LHS: $$\sqrt{2}\,\tan\dfrac{\pi}{8} = \dfrac{\sqrt{2}\,\sin(\pi/8)}{\cos(\pi/8)}$$

RHS: $$\dfrac{1}{2\cos^2(\pi/8)}$$

We check if LHS = RHS:

$$\dfrac{\sqrt{2}\,\sin(\pi/8)}{\cos(\pi/8)} = \dfrac{1}{2\cos^2(\pi/8)}$$

$$2\sqrt{2}\,\sin(\pi/8)\cos(\pi/8) = 1$$

$$\sqrt{2}\,\sin(\pi/4) = \sqrt{2} \cdot \dfrac{\sqrt{2}}{2} = 1 \quad \checkmark$$

The correct answer is Option C: $$\sqrt{2}\,f\left(\dfrac{a}{2}\right) = f'\left(\dfrac{a}{2}\right)$$.