The value of $$$2\sin\left(\frac{\pi}{8}\right)\sin\left(\frac{2\pi}{8}\right)\sin\left(\frac{3\pi}{8}\right)\sin\left(\frac{5\pi}{8}\right)\sin\left(\frac{6\pi}{8}\right)\sin\left(\frac{7\pi}{8}\right)$$$ is:

We have to evaluate

$$2\sin\left(\frac{\pi}{8}\right)\sin\left(\frac{2\pi}{8}\right)\sin\left(\frac{3\pi}{8}\right)\sin\left(\frac{5\pi}{8}\right)\sin\left(\frac{6\pi}{8}\right)\sin\left(\frac{7\pi}{8}\right).$$

$$\sin\!\left(\frac{2\pi}{8}\right)=\sin\!\left(\frac{\pi}{4}\right).$$

$$\sin\!\left(\frac{3\pi}{8}\right)=\sin\!\left(\frac{\pi}{2}-\frac{\pi}{8}\right)=\cos\!\left(\frac{\pi}{8}\right).$$

$$\sin\!\left(\frac{5\pi}{8}\right)=\sin\!\left(\frac{\pi}{2}+\frac{\pi}{8}\right)=\cos\!\left(\frac{\pi}{8}\right).$$

$$\sin\!\left(\frac{6\pi}{8}\right)=\sin\!\left(\frac{\pi}{2}+\frac{\pi}{4}\right)=\cos\!\left(\frac{\pi}{4}\right).$$

$$\sin\!\left(\frac{7\pi}{8}\right)=\sin\!\left(\pi-\frac{\pi}{8}\right)=\sin\!\left(\frac{\pi}{8}\right).$$

Substituting these values, we get

$$2\sin\!\left(\frac{\pi}{8}\right)\cos\!\left(\frac{\pi}{8}\right)\cdot\sin\!\left(\frac{\pi}{4}\right)\cos\!\left(\frac{\pi}{4}\right)\cdot\sin\!\left(\frac{\pi}{8}\right)\cos\!\left(\frac{\pi}{8}\right).$$

Now use the double-angle identity

$$2\sin\theta\cos\theta=\sin2\theta.$$

Hence,

$$2\sin\!\left(\frac{\pi}{8}\right)\cos\!\left(\frac{\pi}{8}\right)=\sin\!\left(\frac{\pi}{4}\right),$$

and

$$\sin\!\left(\frac{\pi}{8}\right)\cos\!\left(\frac{\pi}{8}\right)=\frac12\sin\!\left(\frac{\pi}{4}\right).$$

Therefore,

$$\sin\!\left(\frac{\pi}{4}\right)\cdot\sin\!\left(\frac{\pi}{4}\right)\cdot\cos\!\left(\frac{\pi}{4}\right)\cdot\frac12\sin\!\left(\frac{\pi}{4}\right).$$

Since

$$\sin\!\left(\frac{\pi}{4}\right)=\cos\!\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2},$$

we obtain

$$=\frac12\left(\frac{\sqrt2}{2}\right)^4=\frac12\cdot\frac14=\frac18.$$

Hence, the correct answer is Option B.