The correct order of dipole moments for the given species is

The dipole moment of a molecule depends on its geometry and the vector sum of its individual bond dipoles.

$\text{BF}_3$ has a symmetrical trigonal planar geometry. The three B-F bond dipoles cancel each other out completely, resulting in a net dipole moment of zero ($\mu = 0$).

$\text{NH}_4^+$ has a symmetrical tetrahedral geometry. The four N-H bond dipoles cancel each other out, resulting in a net dipole moment of zero ($\mu = 0$).

Thus, $\mu(\text{BF}_3) = \mu(\text{NH}_4^+) = 0$.

Both $\text{NH}_3$ and $\text{NF}_3$ have a trigonal pyramidal geometry with one lone pair on the central nitrogen atom.

In $\text{NH}_3$, nitrogen is more electronegative than hydrogen. The N-H bond dipoles point towards the nitrogen atom, which is in the same direction as the orbital dipole of the lone pair. These dipoles reinforce each other, giving a higher net dipole moment.

In $\text{NF}_3$, fluorine is more electronegative than nitrogen. The N-F bond dipoles point away from the nitrogen atom, which is in the opposite direction to the orbital dipole of the lone pair. These dipoles partially cancel each other out, resulting in a lower net dipole moment.

Therefore, $\mu(\text{NF}_3) < \mu(\text{NH}_3)$.

The correct order of dipole moments is $\text{BF}_3 = \text{NH}_4^+ < \text{NF}_3 < \text{NH}_3$.

Answer: $\text{BF}_3 = \text{NH}_4^+ < \text{NF}_3 < \text{NH}_3$