Is there a statement independent from PA and does not increase the consistency strength? There are many known independence results of $\mathsf{PA}$, for example, Goodstein theorem, Paris-Harrington theorem, and the reflection principle. But these examples imply the consistency of $\mathsf{PA}$.
I guess not all independent statements of $\mathsf{PA}$ imply the consistency of $\mathsf{PA}$. I have tried to find such examples, but I can't even prove the existence of such a proposition. I would appreciate your help.

Here's a proof that there exists such a sentence (although as Carl observes, this doesn't give an example of such a sentence). We'll assume $PA$ is consistent throughout.
Suppose towards a contradiction that for every $\varphi$ which is independent of $PA$, either $\varphi$ or $\neg\varphi$ proves $Con(PA)$. Note that since $PA$ is consistent, this means that for any $\varphi$ which is independent of $PA$, exactly one of $\varphi$ and $\neg\varphi$ implies $Con(PA)$ (if both $\varphi$ and $\neg\varphi$ implied $Con(PA)$, then $PA\vdash Con(PA)$, so $PA$ is inconsistent).
But this lets us construct a computable consistent completion of $PA$, as follows. Enumerate the sentences of arithmetic as $\{\psi_i: i\in\mathbb{N}\}$, and define a sequence of sentences $\varphi_i$ $(i\in\mathbb{N})$ as follows:


*

*Suppose we've defined $\varphi_j$ for all $j<i$. Let $\theta_i=\bigwedge_{j<i}\varphi_j$ be the conjunction of all the $\varphi$s so far. By induction, we'll have four cases, exactly one of which holds: either $PA+\theta_i$ proves $\psi_i$, or $PA+\theta_i$ proves $\neg\psi_i$, or $PA+\theta_i+\psi_i$ proves $Con(PA)$, or $PA+\theta_i+\neg\psi_i$ proves $Con(PA)$.

*By searching through all possible proofs from $PA+\theta_i$, we may determine effectively which of these four cases holds. We now define $\varphi_i$ as follows. If case $1$ or $4$ holds, $\varphi_i=\psi_i$; if case $2$ or $3$ holds, $\varphi_i=\neg\psi_i$.
It's now not hard to check that $PA\cup\{\varphi_i: i\in\mathbb{N}\}$ is a computable consistent completion of $PA$ - contradicting Godel's theorem.
