1. Identify the molecular geometry of each option using VSEPR theory / known coordination geometry.

2. Option A: $\mathrm{PCl}_5$

• Central atom: $\mathrm{P}$
• Number of bonded pairs around P = 5
• Lone pairs on P = 0

So the steric number is $5$, giving trigonal bipyramidal geometry.

Hence, $\mathrm{PCl}_5$ is not square pyramidal.

3. Option B: $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$

• Oxidation state of Ni: $x + 4(-1) = -2 \Rightarrow x = +2$
• So Ni is $\mathrm{Ni}^{2+}$, i.e. $3d^8$.
• Since $\mathrm{CN}^-$ is a strong field ligand, this complex is typically square planar.

Thus, $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ is square planar, not square pyramidal.

4. Option C: $\mathrm{PF}_5$

• Central atom: $\mathrm{P}$
• Bonded pairs = 5
• Lone pairs = 0

Again, steric number $=5$, so geometry is trigonal bipyramidal.

Hence, $\mathrm{PF}_5$ is not square pyramidal.

5. Option D: $\mathrm{BrF}_5$

• Central atom: $\mathrm{Br}$
• Valence electrons on Br = 7
• It forms 5 bonds with F, leaving 1 lone pair.
• Therefore, total electron pair regions around Br = 6.

So the electron pair geometry is octahedral. With one lone pair, the molecular shape becomes square pyramidal.

Hence, $\mathrm{BrF}_5$ has square pyramidal shape.

6. Conclusion

The molecule / ion with square pyramidal shape is:

$\boxed{\mathrm{BrF}_5}$

So the correct option is D.