Let $A$ be such a group, and let $B_i = \{a\in A\ |\ |a|=p\text{ and } h(a)=i\}$. Note that at least one of the $B_i$ is non-empty by Lemma 8 in section 9. Also, if there existed an $N$ such that for all $m>N$, $B_m$ was empty, then for all $a\in A$, we would have $p^{N+1}a=0$, so $A$ would have bounded order. Thus infinitely many of the $B_i$ are non-empty; now simply mimic the proof of Theorem 9, using Lemma 7 and Theorem 7.
EDIT - Sorry, I left out a couple details, which I don't mind filling in. First, $h(a)$ means the height of $a$. Second, if such an $N$ existed as above, then every element of order $p$ in $p^{N+1}A$ would have infinite height in $A$. It is easy to see this implies it has infinite height in $p^{N+1}A$. Thus $p^{N+1}A$ is divisible; since $A$ is reduced, it is $0$.