The hybridization of orbital of $\mathrm{N}$ atom in ${\mathrm{NO}}_3^{-}, {\mathrm{NO}}_2^{+}$, and ${\mathrm{NH}}_4^{+}$, respectively

 The ion ${\mathrm{NO}}_3$has $24 (= 5 + 3 \mathrm{x} 6 + 1)$ valence electrons. These will be distributed as

There are 3 electron-pairs is around $\mathrm{N}$. The shape of ${\mathrm{NO}}_3$ will be trigonal planar. Hence ${\mathrm{sp}}^2$ hybridization of $\mathrm{N}$ orbitals is involved.

The ion ${\mathrm{NO}}_1$ has $16 ( = 5 + 2 \mathrm{x} 6 - 1)$ valence electrons. These will be distributed as

There are 2 electron pairs around $\mathrm{N}$. The shape of ${\mathrm{NO}}_1$ will be linear. Hence, $\mathrm{sp}$ hybridization of ${\mathrm{NO}}_2^{+}$

The above two hybridization match with the given choice( c ). It can be shown that the ${\mathrm{NH}}_4^{+}$ involves ${\mathrm{sp}}^3$ hybridization of N orbitals. The valence electrons in it is $8 (= 5 + 4 \mathrm{x} 1 - 1 )$. These are distributed as

These are four electron-pairs around $\mathrm{N}$. The shape of ${\mathrm{NH}}_4^{+}$ will be tetrahedral. Hence, ${\mathrm{sp}}^3$ hybridization of N orbitals is involved.