(totally) (M,P)-generic forcing condition We say a cardinal $\theta$ is sufficiently large for a forcing $Q$ if $\mathcal{P}(\mathcal{P}(Q)) \in H(\theta)$.
And a set $M$ is a suitable model for $Q$ if $Q \in M$ and $M \prec H(\theta)$, $M$ countable, for some $\theta$ which is sufficiently large for $Q$.
For a suitable model $M$ for $Q$ and a condition $q \in Q$ we say that $q$ is $(M,Q)$-generic if whenever $r \leqslant q$, $D \in M$ dense, $D \subset Q$, $r$ is compatible with an element of $D \cap M$.
If $\lbrace p \in Q \cap M \colon q \leqslant p \rbrace$ is an $(M,Q)$-generic filter, then $q$ is called totally $(M,Q)$-generic.
$Q$ is totally proper if whenever $M$ is a suitable model for $Q$ and $q \in Q \cap M$, $q$ has a totally $(M,Q)$-generic extension.

Let $Q$ a totally proper forcing and $M$ a suitable model for $Q$.
Then not every $(M,Q)$-generic condition needs to be totally $(M,Q)$-generic.
But is it possible to extend every $(M,Q)$-generic condition to a totally $(M,Q)$-generic condition?

I would think yes but no proof.
 A: I think the following argument works to show that generic conditions do extend to totally generic ones.
As mentioned in the comments, a poset $Q$ is totally proper iff it is proper and countably distributive. I will only need the forward direction and this isn't difficult to see: given a condition $p$ and a name $\dot{f}$ for a countable sequence take a suitable $M\ni p,\dot{f}$. Now, for each $n$, the hull $M$ also contains the dense set of conditions deciding $\dot{f}(n)$. It follows that any totally $(M,Q)$-generic condition must decide $\dot{f}$, and, since we can find one below $p$, it is dense in $Q$ that $\dot{f}$ is in the ground model.
For the main argument let $Q$ be totally proper, $M$ a suitable model and $q$ an $(M,Q)$-generic condition. Enumerate the maximal antichains in $M$ by $\langle A_n;n<\omega\rangle$. For each $n$ let $D_n$ be the set of conditions extending some element of $A_n\cap M$. Then $D_n$ is open dense below $q$, since this condition was $(M,Q)$-generic. Since $Q$ is countably distributive, $\bigcap_n D_n$ is dense below $q$ and any condition in this intersection is totally $(M,Q)$-generic. So, in fact, $q$ extends to many totally $(M,Q)$-generic conditions.
