1. We need to evaluate

$P=\cos\frac{\pi}{2^2}\cdot \cos\frac{\pi}{2^3}\cdots \cos\frac{\pi}{2^{10}}\cdot \sin\frac{\pi}{2^{10}}.$

That is,

$P=\left(\prod_{k=2}^{10}\cos\frac{\pi}{2^k}\right)\sin\frac{\pi}{2^{10}}.$

2. Use the standard identity

$\sin 2x = 2\sin x\cos x$

which gives

$\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}.$

Applying repeatedly,

$\sin x = 2^n \sin\frac{x}{2^n}\cos\frac{x}{2}\cos\frac{x}{2^2}\cdots \cos\frac{x}{2^n}.$

Hence,

$\cos\frac{x}{2}\cos\frac{x}{2^2}\cdots \cos\frac{x}{2^n}\sin\frac{x}{2^n} = \frac{\sin x}{2^n}.$

3. Here, take $x=\dfrac{\pi}{2}$. Then

$\cos\frac{\pi}{2^2}\cos\frac{\pi}{2^3}\cdots \cos\frac{\pi}{2^{10}}\sin\frac{\pi}{2^{10}} = \frac{\sin(\pi/2)}{2^9}.$

Why $2^9$? Because the cosine factors run from $2^2$ to $2^{10}$, i.e. there are $9$ cosine terms.

4. Now,

$\sin\frac{\pi}{2}=1.$

So,

$P=\frac{1}{2^9}=\frac{1}{512}.$

5. Therefore the correct option is

$\boxed{\frac{1}{512}}.$

So, option $\boxed{D}$ is correct.