It will be like -
$for (x)$
{
$if (F(x) \ and \ P(x))$
{
$for(y)$
{
$ if(M(x,y)) $
$break;$
}
}
}
In $\forall x((F(x) \wedge P(x)) \rightarrow \exists y M(x, y)), y$ does not occur as a free variable at LHS of implication so you can move it to left side of whole predicate.
$\forall x((F(x) \wedge P(x)) \rightarrow \exists y M(x, y)) \equiv \forall x \exists y((F(x) \wedge P(x)) \rightarrow M(x, y))$
For Null Quantification - Refer this