Determine the geometry of each species by considering the central atom's electron count, hybridisation, and ligand field strength.

$[Co(CN{)}_4{]}^{4-}$: Co is in $0$ oxidation state, $3d^{7}4s^2$. With $C{N}^{-}$ (strong field ligand), the configuration pairs up and one electron is promoted, giving $ds{p}^2$ hybridisation. Geometry: square planar.

$[Co(CO{)}_3(NO)]$: Co is in $-1$ oxidation state, $3d^{10}$. With strong field $CO$ and $N{O}^{+}$, hybridisation is $s{p}^3$. Geometry: tetrahedral.

$Xe{F}_4$: $4$ bond pairs and $2$ lone pairs, $s{p}^3{d}^2$. Geometry: square planar.

$[PC{l}_4{]}^{+}$: $4$ bond pairs, $0$ lone pairs, $s{p}^3$. Geometry: tetrahedral.

$[PdC{l}_4{]}^{2-}$: Pd${}^{2+}$ is $4d^8$; $C{l}^{-}$ behaves as a strong field ligand for second-row transition metals leading to $ds{p}^2$. Geometry: square planar.

$[IC{l}_4{]}^{-}$: $4$ bond pairs and $2$ lone pairs, $s{p}^3{d}^2$. Geometry: square planar.

$[Cu(CN{)}_4{]}^{3-}$: Cu${}^{+}$ is $3d^{10}$, $s{p}^3$. Geometry: tetrahedral.

$P_4$: white phosphorus has four P atoms at the corners of a regular tetrahedron, but the molecule as a whole has tetrahedral $shape$ rather than tetrahedral geometry around any one central atom. Conventionally in JEE counting questions, $P_4$ is not labelled as having tetrahedral geometry in the central-atom sense; only molecules with one central atom in $s{p}^3$ arrangement (or related) count.

Species with tetrahedral geometry: $[Co(CO{)}_3(NO)]$, $[PC{l}_4{]}^{+}$, $[Cu(CN{)}_4{]}^{3-}$. Total $=3.$