There is a serious typo in Problem 18.6 of Belaniuk's text, because the $\mathcal{A}$ in the problem refers to a (rather weak) fragment of $PA$ (but strong enough to numeralwise represent all recursive functions). His $\mathcal{A}$ is defined on p.117 of his text . Note that by Th($\mathcal{A}$) Belaniuk refers to the deductive closure of $\mathcal{A}$ (which is a rather unusual notation)
1. As stated, the problem is patently false as suspected by Dave. There are two ways to fix it.
2. One is to replace $\mathcal{A}$ by the true theory of arithmetic (often denoted $TA$), as suggested by Noah.
3. Another way to fix the problem, and a more interesting one (since it takes advantage of the fact that $\mathcal{A}$ is "smart enough" to prove all true existential sentences of arithmetic) is to reformulate the problem as follows:
Given $\Sigma $ as in the exercise, show that there is a formula Incon($\Sigma $) such that: $\Sigma $ is inconsistent iff $\mathcal{A}\vdash $Incon($\Sigma $), or equivalently, $\Sigma $ is consistent iff $\mathcal{A}\nvdash $Incon($\Sigma $).