The complex compound [CuII(NH3)4][PtIICl4] consists of two coordination complexes joined together. To determine the total number of possible isomers for this compound, we need to consider the different arrangements of ligands around the central metal ions of each complex.

[CuII(NH3)4] Complex:
- The central metal ion is CuII (copper in the +2 oxidation state).
- It is surrounded by four ammonia (NH3) ligands, which are neutral and uncharged.
- Since ammonia is a monodentate ligand, each NH3 ligand can only form one bond with the metal ion.
- The four NH3 ligands can arrange themselves around the copper ion in different ways, resulting in different isomers.

[PtIICl4] Complex:
- The central metal ion is PtII (platinum in the +2 oxidation state).
- It is surrounded by four chloride (Cl-) ligands, which are also neutral.
- Since chloride is a monodentate ligand, each Cl- ligand can only form one bond with the metal ion.
- The four Cl- ligands can arrange themselves around the platinum ion in different ways, resulting in different isomers.

Total Number of Possible Isomers:
To determine the total number of possible isomers, we multiply the number of isomeric possibilities for each complex together.

- For the [CuII(NH3)4] complex, there are two possible geometries: square planar and tetrahedral. Each geometry has one isomer.
- For the [PtIICl4] complex, there is only one possible geometry: square planar. It has one isomer.

Therefore, the total number of possible isomers for the compound [CuII(NH3)4][PtIICl4] is 1 (isomer from [CuII(NH3)4]) x 1 (isomer from [PtIICl4]) = 1 x 1 = 1.

Hence, the correct answer is '1'.