The molecule in which hybrid MOs involve only one d-orbital of the central atom is:

We begin by recalling that the type of hybridisation tells us how many and which orbitals of the central atom mix to form the hybrid molecular orbitals (MOs).

For our purpose we must identify the species in which exactly one 3d (or 4d, 5d …) orbital of the central atom participates.

The possible common hybrid schemes are:

$$\text{dsp} ^2 :\; 1d + 1s + 2p \;( \text{total } 4$$ hybrids, square-planar $$)$$

$$\text{sp} ^3 \text{d} ^2 \;( \text{or } \text{d} ^2 \text{sp} ^3):\; 2d + 1s + 3p \;( \text{total } 6$$ hybrids, octahedral or its derivatives $$)$$

Thus, a species showing dsp2 hybridisation will involve only one d orbital, whereas any species with sp3d2 (or d2sp3) must necessarily involve two d orbitals.

Now we analyse each option one by one.

Option A: $$[Ni(CN)_4]^{2-}$$

Nickel is the central atom. First we write its ground-state configuration:

$$\mathrm{Ni}:\; [Ar]\,3d^8 4s^2$$

In the complex, nickel is present as $$\mathrm{Ni^{2+}}$$, so we remove two 4s electrons:

$$\mathrm{Ni^{2+}}:\; [Ar]\,3d^8$$

The ligand $$\mathrm{CN^-}$$ is a strong field ligand (spectrochemical series), so it pairs up the 3d electrons. After pairing we have:

$$\text{Filled }3d: \uparrow\downarrow\,\uparrow\downarrow\,\uparrow\downarrow\,\uparrow\downarrow$$
$$\text{Empty }3d: \_\_\_$$

Thus one 3d orbital (usually $$d_{x^2-y^2}$$) becomes vacant and can participate in hybridisation together with one 4s and two 4p orbitals:

$$1(3d) + 1(4s) + 2(4p) \;\longrightarrow\; 4\,\text{dsp}^2\text{ hybrids}$$

This is square-planar geometry. Hence only one d-orbital is involved.

Option B: $$BrF_5$$

$$\mathrm{Br}$$ has seven valence electrons, and five $$\mathrm{F}$$ atoms plus one lone pair give six electron domains in total. Hybridisation is therefore $$\text{sp}^3\text{d}^2$$ (or $$\text{d}^2\text{sp}^3$$), which uses two d orbitals.

Option C: $$XeF_4$$

For xenon, six electron domains (four bonds + two lone pairs) again require $$\text{sp}^3\text{d}^2$$ hybridisation, involving two d orbitals.

Option D: $$[CrF_6]^{3-}$$

Whether we describe it as inner-orbital $$\text{d}^2\text{sp}^3$$ or outer-orbital $$\text{sp}^3\text{d}^2$$, an octahedral species always employs two d orbitals.

From the above comparison, the only species whose hybrid MOs draw upon exactly one d orbital is $$[Ni(CN)_4]^{2-}$$.

Hence, the correct answer is Option A.