Consider $N\left(A\right)\mathit{=}m\text{and}N\left(B\right)\mathit{=}n$

$P_B$ has $112$ elements more than $P_A$

$N\left(P_B\right)=112+N\left(P_A\right)$

$\Rightarrow N\left(P_B\right)-N\left(P_A\right)=112$

$\Rightarrow 2^n-2^m=16\times 7$

Rewrite the above equation.

$2^m\left(2^{n-m}-1\right)=16\times 7$

$\Rightarrow 2^m\left(2^{n-m}-1\right)=2^4\left(2^3-1\right)$

From the above equation

$\begin{array}{r}m=4\end{array}$

$\Rightarrow \begin{array}{r}n-m=3\end{array}$

$\Rightarrow n-4=3$

$\Rightarrow n\mathit{=}7$

The number of injective function from $A$ to $B$

$={}^{N\left(B\right)}P_{N\left(A\right)}$

$={}^{7}P_4$

$=\frac{7!}{3!}$

$=840$