Hybridization of central atom in ${\mathrm{NF}}_3$ is: 

The general formula to find the type of hybridization in a given type of compounds is:

$\mathrm{X}=\frac{1}{2}(\mathrm{V}+\mathrm{M}+\mathrm{A}-\mathrm{C})$ where $\mathrm{V}=$number of electrons

in valence shell; $M =$ number of monovalent atoms;

$\mathrm{A}=$charge on anion and $\mathrm{C}=$ charge on cation. Central atom, has $5($i.e. $2$ of $2\mathrm{s}+3$ of $2\mathrm{p})$ valence electrons;

no. of monovalent atoms in ${\mathrm{NF}}_3=3.$ Thus ;$x=\frac{1}{2}(5+3+0-0)=4.$Hence, type of hybridization is ${\mathrm{sp}}^3(\therefore 1+3=4).$