It can be shown (using Löb's theorem) that $$PA \nvdash ConPA \rightarrow \neg \Box \neg ConPA.$$
But what can be said about this sentence in the standard model of arithmetic $\mathcal{N} = (\mathbb{N}, 0, S, +, \cdot)$?
Do we also have $\mathcal{N} \nvDash ConPA \rightarrow \neg \Box \neg ConPA$?
And as a side question, I'm slightly confused by the assumed consistency of Peano arithmetic. Usually it is assumed that PA is consistent, but obviously by Gödel's second incompleteness theorem this cannot be proven within PA (if it is indeed the case). We also know that $\mathcal{N}$ is a model of PA. However, do we have $\mathcal{N} \vDash ConPA$?
Thanks in advance for any help!
Well (under standard assumptions) we do indeed have $\mathcal{N}\models Con(PA)$; the problem is that there are other models of $PA$ which don't satisfy $Con(PA)$. Indeed, by the Completeness Theorem, the statement "$Con(PA)$ is independent of $PA$" is exactly equivalent to the statement "There are models of $PA$ in which $Con(PA)$ holds, and there are models in which it fails".
As to the sentence "$Con(PA)\implies \neg \Box\neg Con(PA)$" (I assume "$\Box$" means "is provable in $PA$", here), the answer to your question is "no": both the antecedent and the consequent are true in $\mathcal{N}$, so $\mathcal{N}$ does satisfy this sentence. The reason this sentence is not provable in $PA$ is that there can be models of $PA$ in which
$PA$ is consistent, but
$PA+Con(PA)$ is not consistent.