1. Identify the types of isomerism possible

   The complex is square planar: $[Pt(Cl)(NO_2)(NO_3)(SCN)]^{2-}$

   It has four different monodentate ligands:

   • $Cl^-$
   • $NO_2^-$
   • $NO_3^-$
   • $SCN^-$

   Since all four ligands are different, a square-planar complex of type $[Mabcd]$ has: $\frac{4!}{4}=6$ geometrical arrangements? Actually, for square planar complexes with four different ligands, the number of geometrical isomers equals the number of distinct choices of trans pairs.

   The number of ways to divide 4 ligands into 2 trans pairs is: $\frac{1}{2}\binom{4}{2}=3$

   So, ignoring linkage for the moment, there are 3 geometrical isomers.

2. Check ligands capable of linkage isomerism

   Among the ligands:

   • $Cl^-$ : not ambidentate
   • $NO_3^-$ : usually taken as monodentate through O here, no linkage counted in this context
   • $NO_2^-$ : ambidentate, can bind as
     • nitro: $M-NO_2$
     • nitrito: $M-ONO$
   • $SCN^-$ : ambidentate, can bind as
     • thiocyanato: $M-SCN$
     • isothiocyanato: $M-NCS$

   Thus there are 2 independent ambidentate ligands, each with 2 linkage possibilities.

   Hence total linkage possibilities: $2 \times 2 = 4$

3. Total number of isomers

   For each of the 3 square-planar geometrical arrangements, there are 4 linkage variants.

   Therefore total isomers: $3 \times 4 = 12$

4. Check options

   • A: 8 ❌
   • B: 12 ✅
   • C: 16 ❌
   • D: 24 ❌

5. Final answer

   $\boxed{12}$

So the correct option is B.