By $\textbf{PA}$ I will mean the usual first-order Peano Arithmetic. I will denote an element of $\mathbb{N}$ by $n$, and by $[n]$ I will denote the corresponding term in the language of $\textbf{PA}$: $\underbrace{1+1+\dotsb+1}_{\text{$n$ times}}$.
Let $P(x)$ be a recusive predicate defined in $\textbf{PA}$, and let $\theta$ be the sentence $(\forall x)(P(x))$.
Suppose that this sentence $\theta$ is independent from $\textbf{PA}$, that is, $\textbf{PA} \nvdash \theta$ and $\textbf{PA} \nvdash \neg \theta$. Then it follows that $\mathbb{N} \models \theta$. For otherwise, we would have $\mathbb{N} \models \neg \theta$, and hence, for some $n \in \mathbb{N}$, we would have $\mathbb{N} \models \neg P(n)$. Since $n$ is a standard natural number and $P(x)$ is recursive, this would imply that $\textbf{PA} \vdash \neg P([n])$ — contradicting the independence of $\theta$.
An often discussed example of the above situation is the following one: if Goldbach's Conjecture is independent of $\textbf{PA}$, then it is true (that is, it holds in the standard model $\mathbb{N}$).
My question is the following:
For what kind of predicates $P(x)$ definable in $\textbf{PA}$ does the independence of the sentence $(\forall x)(P(x))$ from $\textbf{PA}$ imply that $\mathbb{N} \models (\forall x)(P(x))$?
Note: I have only a modest knowledge of mathematical logic, so I apologize in advance if this question turns out to have a trivial answer.
