Method 1 :-
According to remainder theorem, the remainder for $\frac{f\left(x\right)}{x+a}$ is f(-a).
Let the number to be subtracted be 'k'.
Since, x+3 divides $x^3+4x^2-7x+12-k$ perfectly $⟹$ the remainder when x is substituted by -3 is 0$⟹f(-3)=0$.
$\Rightarrow (-3{)}^3+4(-3{)}^2-7(-3)+12-k=0$
$\Rightarrow -27+36+21+12=k$
$\Rightarrow k=42$
Method 2 :-
If we want $x^3+4x^2-7x+12$ to be perfectly divisible by x+3,
$x^3+4x^2-7x+12$ should be of the form of (x+3)$\times$g(x).
$(x^3+4x^2-7x+12)$ 
$=(x+3)(x^2+x-10)+12+30$
$=(x+3)(x^2+x-10)+42$
So, the number that needs to be subtracted = 42.
Method 3 :- Shortcut!
Since this is a variable-number question, assume x=1; then the expression P=10.
Now we need to check which answer option satisfies the following condition.
$\frac{10-\left(answeroption\right)}{4_{\left(Remainder\right)}}=0$
${\frac{10-42}{4}}_{\left(Remainder\right)}=-{\frac{32}{4}}_{\left(Remainder\right)}=0$