If $\operatorname{Con}(\mathrm{PA})$, then $\operatorname{Con}(\mathrm{PA}+\operatorname{Con}(\mathrm{PA}))$? Assume $\newcommand\PA{\mathrm{PA}}\newcommand\Con{\operatorname{Con}}$ that $\PA$ is consistent. Then we know that $\PA$ cannot prove $\Con(\PA)$. I was wondering. Can $\PA$ prove that $$\Con(\PA) \Rightarrow \Con(\PA + \Con(\PA))?$$

I assume that you mean can we prove from $\
ewcommand\PA{\mathsf{PA}}\
ewcommand\Con{\operatorname{Con}}\PA+\Con(\PA)$ the statement $\Con(\PA+\Con(\PA))$.

Th answer is no. We only added one axiom to $\PA$ so our theory is still recursively enumerable, so Gödel's theorem tells us it cannot prove its own consistency.

Note that $$\PA\vdash\Con(\PA)\rightarrow\Con(\PA+\Con(\PA))\iff\PA+\Con(\PA)\vdash\Con(\PA+\Con(\PA)),$$ so the above argument shows that indeed the implication is not provable in $\PA$.