1. Let $P=2\sin\left(\frac{\pi}{22}\right)\sin\left(\frac{3\pi}{22}\right)\sin\left(\frac{5\pi}{22}\right)\sin\left(\frac{7\pi}{22}\right)\sin\left(\frac{9\pi}{22}\right).$

We need to evaluate this product.

2. Use the standard identity $\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}.$

For $n=11$, $\prod_{k=1}^{10}\sin\left(\frac{k\pi}{11}\right)=\frac{11}{2^{10}}.$

3. Now use symmetry: $\sin\left(\frac{k\pi}{11}\right)=\sin\left(\pi-\frac{k\pi}{11}\right)=\sin\left(\frac{(11-k)\pi}{11}\right).$

Hence, $\prod_{k=1}^{10}\sin\left(\frac{k\pi}{11}\right)=\left[\sin\left(\frac{\pi}{11}\right)\sin\left(\frac{2\pi}{11}\right)\sin\left(\frac{3\pi}{11}\right)\sin\left(\frac{4\pi}{11}\right)\sin\left(\frac{5\pi}{11}\right)\right]^2.$

Therefore, $\sin\left(\frac{\pi}{11}\right)\sin\left(\frac{2\pi}{11}\right)\sin\left(\frac{3\pi}{11}\right)\sin\left(\frac{4\pi}{11}\right)\sin\left(\frac{5\pi}{11}\right)=\frac{\sqrt{11}}{2^5}.$

4. Our given angles are $\frac{\pi}{22},\frac{3\pi}{22},\frac{5\pi}{22},\frac{7\pi}{22},\frac{9\pi}{22},$ which are exactly $\frac{(2r-1)\pi}{22},\quad r=1,2,3,4,5.$

So we use the identity for product of sines of odd multiples: $\prod_{r=1}^{m}\sin\left(\frac{(2r-1)\pi}{4m+2}\right)=\frac{1}{2^m}.$

Here $4m+2=22$, so $m=5$. Thus \sin\left(\frac{\pi}{22}\right)\sin\left(\frac{3\pi}{22}\right)\sin\left(\frac{5\pi}{22}\right)\sin\left(\frac{7\pi}{22}\right)\sin\left(\frac{9\pi}{22}\right)=\frac{1}{2^5}=rac{1}{32}.

Hence $P=2\cdot \frac{1}{32}=\frac{1}{16}.$

5. Therefore the correct option is $\boxed{\frac{1}{16}}$ which is option $\boxed{\text{B}}$.

6. Verification with stored answer:

• Derived answer: $\boxed{\text{B}}$
• Stored correct answer: $\boxed{\text{B}}$

They agree.