We check the geometry of each given species and count those which are square planar.

Analyze one by one:

(i) $\mathrm{XeF}_4$

• Central atom Xe has $4$ bond pairs and $2$ lone pairs.
• Electron pair geometry: octahedral
• Molecular shape: square planar
• Hence, this is counted.

(ii) $\mathrm{SF}_4$

• Central atom S has $4$ bond pairs and $1$ lone pair.
• Electron pair geometry: trigonal bipyramidal
• Molecular shape: see-saw
• Not square planar.

(iii) $\mathrm{SiF}_4$

• Central atom Si has $4$ bond pairs and no lone pair.
• Shape: tetrahedral
• Not square planar.

(iv) $\mathrm{BF}_4^-$

• Central atom B has $4$ bond pairs and no lone pair.
• Shape: tetrahedral
• Not square planar.

(v) $\mathrm{BrF}_4^-$

• Total valence electrons around Br correspond to $4$ bond pairs and $2$ lone pairs.
• Steric number $= 6$.
• Electron pair geometry: octahedral
• Molecular shape: square planar
• Hence, this is counted.

(vi) $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$

• $\mathrm{Cu}^{2+}$ is $3d^9$.
• Four-coordinate $\mathrm{Cu}^{2+}$ complexes commonly adopt square planar geometry due to Jahn–Teller distortion / dsp$^2$ arrangement in such standard classification.
• Hence, this is counted.

(vii) $[\mathrm{FeCl}_4]^{2-}$

• Oxidation state of Fe: $x + 4(-1) = -2 \Rightarrow x=+2$
• So Fe is $\mathrm{Fe}^{2+}$, i.e. $3d^6$.
• With weak ligand $\mathrm{Cl}^-$ and coordination number $4$, geometry is typically tetrahedral.
• Not square planar.

(viii) $[\mathrm{PtCl}_4]^{2-}$

• Oxidation state of Pt: $x + 4(-1) = -2 \Rightarrow x=+2$
• So Pt is $\mathrm{Pt}^{2+}$, a $d^8$ system.
• $d^8$ complexes of Pt(II) are characteristically square planar.
• Hence, this is counted.

Therefore, square planar species are:

• $\mathrm{XeF}_4$
• $\mathrm{BrF}_4^-$
• $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$
• $[\mathrm{PtCl}_4]^{2-}$

Total number: $4$