I want to show that there is a $\mathbb P$-name $\sigma$ such that for every $G$ a generic filter we will have $$M[G]\vDash \exists x\phi (x) \iff M[G]\vDash \phi (\sigma [G])$$ while $\phi (x)$ is a formula in the language of forcing, and $x$ is the only free variable in $\phi$. and $M$ is a countable transitive model of ZFC.
I showed a lemma that given an anti-chain $A\subseteq\mathbb P,A\in M$ if we assume that for every $p\in A$ we assign a $\mathbb P$-name $x_p$ the we can find a $\mathbb P$-name $y$ such that for every filter $G\subseteq \mathbb P$ we have that $y[G]=x_p[G]$ if $p\in G\cap A$ (there will always be a single $p\in G\cap A$ if the intersection is not empty) and if not then $y[G]=\emptyset$.
so what i generally tried to do is to take a maximal anti-chain $A$ and then to show that for every $p\in A$ for all generic filters such that $p\in G$ we have a single $\mathbb P$-name $x_p$ such that $$M[G]\vDash \exists x\phi (x) \iff M[G]\vDash \phi (x_p [G])$$ and then by using the lemma we can fined the $\mathbb P$-name $\sigma$ that we need.
More specifically tried to do so by taking an arbitrary maximal anti-chain $A$ and then for every $p\in A$ i will first look at some generic filter $G$ such that $p\in G$. now we have that $$M[G]\vDash \exists x\phi (x) \iff M[G]\vDash \phi (\tau [G])$$ for some $\mathbb P$-name $\tau$. then by the Truth Lemma there exists a $q\in G$ such that $q\Vdash \phi(\tau)$ and by the Extention Lemma there exist $r\in G$ with $r\le p,q$ and $r\Vdash \phi (\tau)$. thus by that way we can construct a anti-chain of thous $r$'s but i think it's not maximal. if it was so we it would have being done. help!
HINT: The set $\{p\in\Bbb P\mid \text{There exists }\dot x_p\text{ such that }p\Vdash\phi(\dot x_p),\text{or }p\Vdash\lnot\exists x\phi(x)\}$ is dense and open, therefore it contains a maximal antichain. Use that one.