Let $$S = \left\{\theta \in \left(0, \frac{\pi}{2}\right) : \sum_{m=1}^{9} \sec\left(\theta + (m-1)\frac{\pi}{6}\right) \sec\left(\theta + \frac{m\pi}{6}\right) = -\frac{8}{\sqrt{3}}\right\}$$. Then

Since consecutive angles differ by $$\dfrac{\pi}{6}$$, we use the identity $$\sec A \cdot \sec B = \frac{1}{\cos A \cos B}$$ along with $$B - A = \dfrac{\pi}{6}$$ and $$\sin\!\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$$ to obtain $$\frac{\sin(B-A)}{\cos A \cos B} = \tan B - \tan A \implies \frac{1}{\cos A \cos B} = \frac{\tan B - \tan A}{\sin(B-A)} = 2(\tan B - \tan A).$$

Let $$\theta_m = \theta + (m-1)\dfrac{\pi}{6}$$. Then the given sum becomes $$\sum_{m=1}^{9} \sec\!\left(\theta + (m-1)\frac{\pi}{6}\right)\sec\!\left(\theta + \frac{m\pi}{6}\right) = 2\sum_{m=1}^{9}\left[\tan\!\left(\theta + \frac{m\pi}{6}\right) - \tan\!\left(\theta + \frac{(m-1)\pi}{6}\right)\right],$$ which telescopes.

After cancellation, only the first and last terms remain: $$2\left[\tan\!\left(\theta + \frac{9\pi}{6}\right) - \tan\theta\right] = 2\left[\tan\!\left(\theta + \frac{3\pi}{2}\right) - \tan\theta\right].$$

Using the periodicity of tangent ($$\tan$$ has period $$\pi$$) gives $$\tan\!\left(\theta + \frac{3\pi}{2}\right) = \tan\!\left(\theta + \frac{3\pi}{2} - \pi\right) = \tan\!\left(\theta + \frac{\pi}{2}\right) = -\cot\theta.$$ Hence the sum becomes $$2(-\cot\theta - \tan\theta) = -2\left(\frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta}\right) = -2 \cdot \frac{\cos^2\theta + \sin^2\theta}{\sin\theta\cos\theta} = \frac{-2}{\sin\theta\cos\theta}.$$

Since $$\sin 2\theta = 2\sin\theta\cos\theta$$, this simplifies to $$\frac{-4}{\sin 2\theta}.$$

Setting this equal to $$\frac{-8}{\sqrt{3}}$$ yields $$\frac{-4}{\sin 2\theta} = \frac{-8}{\sqrt{3}} \implies \sin 2\theta = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}.$$ For $$\theta \in \left(0, \dfrac{\pi}{2}\right)$$, we have $$2\theta \in (0, \pi)$$, so $$2\theta = \frac{\pi}{3} \implies \theta = \frac{\pi}{6} \quad \text{or} \quad 2\theta = \frac{2\pi}{3} \implies \theta = \frac{\pi}{3},$$ giving $$S = \left\{\dfrac{\pi}{6},\ \dfrac{\pi}{3}\right\}$$.

Finally, $$\sum_{\theta \in S} \theta = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi + 2\pi}{6} = \boxed{\frac{\pi}{2}}.$$ The answer is Option C.