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

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

First we notice the trigonometric fact $$\sin(\theta)=\sin(\pi-\theta).$$ Using this fact, three of the six sines can be rewritten:

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

So the entire product pairs up neatly:

$$\sin\!\left(\frac{\pi}{8}\right)\sin\!\left(\frac{5\pi}{8}\right)=\sin^2\!\left(\frac{\pi}{8}\right),$$ $$\sin\!\left(\frac{2\pi}{8}\right)\sin\!\left(\frac{6\pi}{8}\right)=\sin^2\!\left(\frac{\pi}{4}\right),$$ $$\sin\!\left(\frac{3\pi}{8}\right)\sin\!\left(\frac{7\pi}{8}\right)=\sin^2\!\left(\frac{3\pi}{8}\right).$$

Hence the whole expression becomes

$$$2\;\times\;\sin^2\!\left(\frac{\pi}{8}\right)\;\times\;\sin^2\!\left(\frac{\pi}{4}\right)\;\times\;\sin^2\!\left(\frac{3\pi}{8}\right).$$$

We already know $$\sin\!\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2},$$ so

$$\sin^2\!\left(\frac{\pi}{4}\right)=\left(\frac{\sqrt{2}}{2}\right)^2=\frac12.$$

Now we compute $$\sin\!\left(\frac{\pi}{8}\right)$$ and $$\sin\!\left(\frac{3\pi}{8}\right)$$ using the half-angle formula $$\sin^2\!\left(\frac{\alpha}{2}\right)=\frac{1-\cos\alpha}{2}.$$

Taking $$\alpha=\frac{\pi}{4},$$ we get

$$\sin^2\!\left(\frac{\pi}{8}\right)=\frac{1-\cos\!\left(\frac{\pi}{4}\right)}{2}=\frac{1-\frac{\sqrt{2}}{2}}{2}.$$

Taking $$\alpha=\frac{3\pi}{4},$$ we get

$$$\sin^2\!\left(\frac{3\pi}{8}\right)=\frac{1-\cos\!\left(\frac{3\pi}{4}\right)}{2}=\frac{1-\left(-\frac{\sqrt{2}}{2}\right)}{2}=\frac{1+\frac{\sqrt{2}}{2}}{2}.$$$

Let us set

$$$A=\sin^2\!\left(\frac{\pi}{8}\right)=\frac{1-\frac{\sqrt{2}}{2}}{2}, \qquad B=\sin^2\!\left(\frac{3\pi}{8}\right)=\frac{1+\frac{\sqrt{2}}{2}}{2}.$$$

The product of these two is

$$$AB=\frac{1-\frac{\sqrt{2}}{2}}{2}\;\times\;\frac{1+\frac{\sqrt{2}}{2}}{2}=\frac{\left(1-\frac{\sqrt{2}}{2}\right)\!\left(1+\frac{\sqrt{2}}{2}\right)}{4}.$$$

Using $$\left(x-y\right)\left(x+y\right)=x^2-y^2,$$ take $$x=1,\;y=\frac{\sqrt{2}}{2}.$$ Thus the numerator becomes

$$1^2-\left(\frac{\sqrt{2}}{2}\right)^2=1-\frac{2}{4}=1-\frac12=\frac12.$$

Therefore

$$AB=\frac{\frac12}{4}=\frac18.$$

Now multiply by $$\sin^2\!\left(\frac{\pi}{4}\right)=\frac12:$$

$$A\,B\,\sin^2\!\left(\frac{\pi}{4}\right)=\frac18\times\frac12=\frac1{16}.$$

This is the value of the six-sine product without the leading 2. Finally, include that factor 2:

$$2\times\frac1{16}=\frac18.$$

Hence, the correct answer is Option B.