Pullback a model in generic extension to a model in ground Let $M\prec H_\theta$ be countable and $\mathbb P\in M$, then clearly $M[G]\prec H_\theta[G]$ for every $\mathbb P$-generic filter $G$ over $V$. then actually there exists a $\mathbb P$-name $\dot N$ such that $\dot N[G]=M[G]$, roughly speaking $\dot N=M\cap V^{\mathbb P}$($V^{\mathbb P}$ is the class of $\mathbb P$-names). Here by $M[G]$, I mean $\{\dot\sigma[G]:\dot\sigma\in M~ is~a~\mathbb P-name\}$. Is the converse true? More precisely:

Let $N\prec H_\theta[G]$ be countable is there a model $M\prec H_\theta^{V}$ including $\mathbb P$, such that $M[G]=N$?

 A: First, let me point out that you don't need any properness at all
to find the model $M$, if you are willing to find it in the extension in which $N$ lives. Namely, assume $\theta$ is sufficiently reflective and that $N\prec
H_\theta[G]$. Now, simply let $M=N\cap H_\theta$. One can verify that $M\prec H_\theta$ by noting that $H_\theta$ is definable in $H_\theta[G]$ by the ground-model definability theorem, and from this it follows that $M\prec H_\theta$, since it will have the Skolem witnesses that suffice to verify the Tarski-Vaught criterion. Notice also that $M[G]=N$ and therefore 
$M[G]\cap H_\theta=M$. This implies that forcing with $G$ over $M$
did not add any new ordinals, and therefore the usual arguments show that $G$ is $M$-generic. So $M$ is just the kind of model you seek.
The trick here is that we do not get $M$ in $V$, but only possibly
in $V[G]$.
But perhaps you have in mind the requirement that $M\in V$ in your
question, and in this case, the answer is no, you cannot always
achieve this, if you are allowed to select $N$ in $V[G]$.
Consider the case of adding a Cohen real $c$. Pick some very big
$\theta$ and consider $H_\theta$. Pick some countable $M\prec
H_\theta$ that does not exist in $V$. But since the forcing is
c.c.c., this elementary substructure lifts to the case $M[c]\prec
H_\theta[c]$. So $N=M[c]$ is a case where the restriction to the
ground model is not available in $V$. 
