The total number of all possible isomers for the square planar...

The given complex is $\text{K}[\text{M(NCS)(NO}_2\text{)(gly)}]$. The complex anion is $[\text{M(NCS)(NO}_2\text{)(gly)}]^{-}$.

The ligands present in the complex are:
1. $\text{gly}^-$ (glycinate): An unsymmetrical bidentate ligand coordinating through nitrogen (N) and oxygen (O). Let this be represented as $(AB)$.
2. $\text{NCS}^-$: An ambidentate monodentate ligand that can coordinate through nitrogen ($\text{-NCS}$) or sulfur ($\text{-SCN}$).
3. $\text{NO}_2^-$: An ambidentate monodentate ligand that can coordinate through nitrogen ($\text{-NO}_2$) or oxygen ($\text{-ONO}$).

First, we find the number of linkage isomers due to the ambidentate ligands:
$\text{NCS}^-$ has 2 linkage modes (N or S).
$\text{NO}_2^-$ has 2 linkage modes (N or O).
Number of linkage combinations = $2 \times 2 = 4$.

For each linkage combination, the complex is of the type $[\text{M}(a)(b)(AB)]$, where $a$ and $b$ are monodentate ligands and $AB$ is an unsymmetrical bidentate ligand.
In a square planar geometry, the bidentate ligand $(AB)$ must occupy adjacent (cis) positions. The monodentate ligands $a$ and $b$ will occupy the remaining two positions.
This gives rise to 2 geometrical isomers for each linkage combination:
1. Ligand $a$ is trans to A (and $b$ is trans to B).
2. Ligand $a$ is trans to B (and $b$ is trans to A).

Since square planar complexes of this type possess a molecular plane of symmetry, they do not exhibit optical isomerism.

Total number of isomers = (Number of linkage combinations) $\times$ (Number of geometrical isomers per combination)
Total isomers = $4 \times 2 = 8$.

Answer: $8$