Considering the principal values of the inverse trigonometric functions, $$\sin^{-1}\left(\dfrac{\sqrt{3}}{2} x + \dfrac{1}{2}\sqrt{1 - x^2}\right)$$, $$-\dfrac{1}{2} \lt x \lt \dfrac{1}{\sqrt{2}}$$, is equal to

Write the numerical coefficients in the given expression with their trigonometric values:
$$\frac{\sqrt3}{2} = \cos\frac{\pi}{6}, \qquad \frac12 = \sin\frac{\pi}{6}$$

Let $$\alpha = \sin^{-1}x$$. By definition of the principal branch,
$$\alpha \in \left[-\frac{\pi}{2},\,\frac{\pi}{2}\right]$$ and $$\sin\alpha = x$$.

Because $$\alpha$$ lies between $$-\dfrac{\pi}{2}$$ and $$\dfrac{\pi}{2}$$, its cosine is non-negative, hence
$$\cos\alpha = \sqrt{1 - \sin^2\alpha} = \sqrt{1 - x^2}.$$\

Rewrite the inside of $$\sin^{-1}(\; \cdot \;)$$ using $$\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$$:

$$\frac{\sqrt3}{2}\,x + \frac12\sqrt{1 - x^2} \;=\; \sin\alpha\cos\frac{\pi}{6} + \cos\alpha\sin\frac{\pi}{6} \;=\; \sin\!\left(\alpha + \frac{\pi}{6}\right).$$

Therefore
$$\sin^{-1}\!\left(\frac{\sqrt3}{2}\,x + \frac12\sqrt{1 - x^2}\right) = \sin^{-1}\!\left(\sin\!\bigl(\alpha + \tfrac{\pi}{6}\bigr)\right).$$

We must now check whether $$\alpha + \dfrac{\pi}{6}$$ itself lies in the principal range $$\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right]$$ so that $$\sin^{-1}(\sin\theta)=\theta$$ holds unaltered.

Given $$-\dfrac12 \lt x \lt \dfrac1{\sqrt2}\;,$$ we have
$$-\frac{\pi}{6} \lt \alpha = \sin^{-1}x \lt \frac{\pi}{4}.$$

Add $$\dfrac{\pi}{6}$$ throughout:
$$0 \lt \alpha + \frac{\pi}{6} \lt \frac{\pi}{4} + \frac{\pi}{6} = \frac{5\pi}{12}.$$

The interval $$\left(0,\dfrac{5\pi}{12}\right)$$ is completely within the principal interval $$\left[-\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right]$$. Hence

$$\sin^{-1}\!\left(\sin\!\bigl(\alpha + \tfrac{\pi}{6}\bigr)\right) = \alpha + \frac{\pi}{6}.$$

Finally substitute back $$\alpha = \sin^{-1}x$$:

$$\sin^{-1}\!\left(\frac{\sqrt3}{2}\,x + \frac12\sqrt{1 - x^2}\right) = \frac{\pi}{6} + \sin^{-1}x.$$

Thus the expression equals $$\dfrac{\pi}{6} + \sin^{-1}x$$, which matches Option B.