Given a regular triangular pyramid $P-ABC$ with a base side length of $\sqrt{2}$, and $PA$, $PB$, $PC$ are mutually perpendicular, we are to find the volume of the sphere on which all vertices lie.

First, let's find the length of $PA$ using the given information. Since the base is an equilateral triangle and $PA$, $PB$, and $PC$ are mutually perpendicular, we can deduce that $PA$ is a height in the equilateral triangle $ABC$. Thus, applying Pythagoras theorem in triangle $PAB$, we have:
$$PA^2 + PB^2 = AB^2.$$
Given that $AB = \sqrt{2}$ and $PA = PB$ due to the symmetry of the pyramid, we can rewrite this as:
$$2PA^2 = (\sqrt{2})^2.$$
Solving for $PA$ gives:
$$2PA^2 = 2 \implies PA^2 = 1 \implies PA = 1.$$

Next, we use the fact that the vertices of the pyramid lie on a sphere. The distance from $P$ to the center of the base, which lies on the sphere's surface, is equal to the radius $R$ of the sphere. Since $PA$, $PB$, and $PC$ form a right-angled triangle with the sphere's diameter as the hypotenuse, we can relate the diameter $(2R)$ to $PA$:
$$(2R)^2 = PA^2 + PB^2 + PC^2.$$
Given $PA = PB = PC = 1$, this simplifies to:
$$(2R)^2 = 3 \times 1^2 \implies 4R^2 = 3 \implies R^2 = \frac{3}{4} \implies R = \frac{\sqrt{3}}{2}.$$

Finally, we calculate the volume $V$ of the sphere using its radius $R$:
$$V = \frac{4\pi}{3}R^3 = \frac{4\pi}{3} \left(\frac{\sqrt{3}}{2}\right)^3 = \frac{4\pi}{3} \cdot \frac{3\sqrt{3}}{8} = \frac{\sqrt{3}\pi}{2}.$$

Therefore, the volume of the sphere is $\boxed{\frac{\sqrt{3}\pi}{2}}$, which corresponds to option B.