The coordination compound that will give four isomers is:

Let us determine the number of isomers for each given complex:

1. $[Fe(en)_3]Cl_3$: It is an octahedral complex of type $[M(AA)_3]^{n+}$. It lacks any plane or centre of symmetry and therefore exhibits optical isomerism, existing as a pair of enantiomers ($d$ and $l$ forms). Total isomers = 2.
2. $[Co(en)_2Cl_2]Cl$: It is of the type $[M(AA)_2b_2]^{n+}$. It exhibits geometrical isomerism forming cis and trans isomers. The trans-isomer is optically inactive due to the presence of a plane of symmetry, while the cis-isomer is optically active and exists as $d$ and $l$ enantiomers. Total stereoisomers = 3.
3. $[Fe(PPh_3)_3NH_3ClBr]Cl$: It is an octahedral complex of type $[Ma_3bcd]^{n+}$. It exhibits geometrical isomerism and forms exactly 4 geometrical isomers. These consist of 1 facial (fac) isomer (where the three identical ligands occupy the corners of a single octahedral face) and 3 meridional (mer) isomers. The three mer isomers arise because the third '$a$' ligand can be placed trans to ligand '$b$', '$c$', or '$d$' respectively. Total geometrical isomers = 4.
4. $[Co(PPh_3)_3Cl]Cl_3$: It is of the type $[Ma_3b]^{n+}$ and does not show stereoisomerism. Total isomers = 1.

Therefore, the compound $[Fe(PPh_3)_3NH_3ClBr]Cl$ will give four geometrical isomers.