If $$L = \sin^2\left(\frac{\pi}{16}\right) - \sin^2\left(\frac{\pi}{8}\right)$$ and $$M = \cos^2\left(\frac{\pi}{16}\right) - \sin^2\left(\frac{\pi}{8}\right)$$

We have two expressions, namely $$L = \sin^{2}\!\left(\dfrac{\pi}{16}\right) - \sin^{2}\!\left(\dfrac{\pi}{8}\right)$$ and $$M = \cos^{2}\!\left(\dfrac{\pi}{16}\right) - \sin^{2}\!\left(\dfrac{\pi}{8}\right).$$ Our task is to simplify each of them and then match the results with the options given.

First we handle $$L$$. We recall the double-angle form for squares of sine: the identity $$\sin^{2}\theta = \dfrac{1 - \cos 2\theta}{2}.$$ Applying this identity to both terms we get

$$\sin^{2}\!\left(\dfrac{\pi}{16}\right) = \dfrac{1 - \cos\!\left(\dfrac{\pi}{8}\right)}{2},$$

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

Substituting these forms into $$L$$ we obtain

$$L = \dfrac{1 - \cos\!\left(\dfrac{\pi}{8}\right)}{2} \;-\; \dfrac{1 - \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

The two fractions have the same denominator, so we can combine the numerators directly:

$$L = \dfrac{1 - \cos\!\left(\dfrac{\pi}{8}\right) - 1 + \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

Inside the numerator the two “1” terms cancel, giving

$$L = \dfrac{-\cos\!\left(\dfrac{\pi}{8}\right) + \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

We know that $$\cos\!\left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}.$$ Substituting this numerical value we arrive at

$$L = \dfrac{-\cos\!\left(\dfrac{\pi}{8}\right) + \dfrac{1}{\sqrt{2}}}{2} = -\dfrac{1}{2}\cos\!\left(\dfrac{\pi}{8}\right) + \dfrac{1}{2\sqrt{2}}.$$

So we have simplified $$L$$ to the form $$L = -\dfrac{1}{2}\cos\!\left(\dfrac{\pi}{8}\right) + \dfrac{1}{2\sqrt{2}}.$$ We keep this result in mind when we compare with the listed choices.

Now we simplify $$M$$. For this we use the companion identity for cosine squares, namely $$\cos^{2}\theta = \dfrac{1 + \cos 2\theta}{2}.$$ Applying it to the first term of $$M$$ yields

$$\cos^{2}\!\left(\dfrac{\pi}{16}\right) = \dfrac{1 + \cos\!\left(\dfrac{\pi}{8}\right)}{2}.$$

The second term $$\sin^{2}\!\left(\dfrac{\pi}{8}\right)$$ we have already expressed above, viz.

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

Substituting both of these into the definition of $$M$$ we get

$$M = \dfrac{1 + \cos\!\left(\dfrac{\pi}{8}\right)}{2} \;-\; \dfrac{1 - \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

Combining the numerators over the common denominator gives

$$M = \dfrac{1 + \cos\!\left(\dfrac{\pi}{8}\right) - 1 + \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

The two “1” terms again cancel, leaving

$$M = \dfrac{\cos\!\left(\dfrac{\pi}{8}\right) + \cos\!\left(\dfrac{\pi}{4}\right)}{2}.$$

Substituting $$\cos\!\left(\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}$$ we obtain

$$M = \dfrac{\cos\!\left(\dfrac{\pi}{8}\right) + \dfrac{1}{\sqrt{2}}}{2} = \dfrac{1}{2}\cos\!\left(\dfrac{\pi}{8}\right) + \dfrac{1}{2\sqrt{2}}.$$

We now compare the simplified forms with the four options furnished in the question.

The expression derived for $$L$$ is $$-\dfrac{1}{2}\cos\!\left(\dfrac{\pi}{8}\right) + \dfrac{1}{2\sqrt{2}},$$ which does not match either Option A or Option B. On the other hand, the expression derived for $$M$$ is $$\dfrac{1}{2\sqrt{2}} + \dfrac{1}{2}\cos\!\left(\dfrac{\pi}{8}\right),$$ and this coincides exactly with Option D.

Hence, the correct answer is Option D.