${\left[{\mathrm{CoCl}}_2{\left({\mathrm{NH}}_3\right)}_4\right]}^{+}+{\mathrm{Cl}}^{-}⟶\left[{\mathrm{CoCl}}_3{\left({\mathrm{NH}}_3\right)}_3\right]+{\mathrm{NH}}_3$

If in the above reaction only one isomer of the product is obtained, which is true for the initial (reactant) complex :

If the reactant $\left[CoC{l}_2\right(N{H}_3{)}_4{]}^{+}$ is in the cis form, the complex $[{\mathrm{CoCl}}_2{\left({\mathrm{NH}}_3\right)}_3{]}^{+}$ will have both a cis (or fac or facial) and a trans (or mer or meridional) isomer.
Only if the reactant $\left[CoC{l}_2\right(N{H}_3{)}_4{]}^{+}$ is in the trans form will the compound$[{\mathrm{CoCl}}_2{\left({\mathrm{NH}}_3\right)}_3{]}^{+}$ have the trans (or mer or meridional) isomer.
Because it can be divided into two isomers, the reactant is in the cis form.

One and alone because all are "b"