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

Simplify the angles: $P=2\sin\left(\frac{\pi}{8}\right)\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{3\pi}{8}\right)\sin\left(\frac{5\pi}{8}\right)\sin\left(\frac{3\pi}{4}\right)\sin\left(\frac{7\pi}{8}\right).$

Use the identity $\sin(\pi-\theta)=\sin\theta.$ Hence, $\sin\left(\frac{7\pi}{8}\right)=\sin\left(\frac\pi8\right),\quad \sin\left(\frac{5\pi}{8}\right)=\sin\left(\frac{3\pi}{8}\right),\quad \sin\left(\frac{3\pi}{4}\right)=\sin\left(\frac\pi4\right).$ So, $P=2\left[\sin\left(\frac\pi8\right)\sin\left(\frac\pi4\right)\sin\left(\frac{3\pi}{8}\right)\right]^2.$

Now use the standard identity $\sin\left(\frac\pi8\right)\sin\left(\frac{3\pi}{8}\right)=\frac{1}{2\sqrt2}.$ Also, $\sin\left(\frac\pi4\right)=\frac{1}{\sqrt2}.$ Therefore, $\sin\left(\frac\pi8\right)\sin\left(\frac\pi4\right)\sin\left(\frac{3\pi}{8}\right)=\frac{1}{2\sqrt2}\cdot \frac{1}{\sqrt2}=\frac14.$

Hence, $P=2\left(\frac14\right)^2=2\cdot \frac{1}{16}=\frac18.$

Therefore the correct option is $\boxed{\frac18}$ which is option C.