Solution:
Consider the set $A' = A \cup \{0\}$ instead of $A$. Let $B = \{a_1, a_2, \ldots, a_k\}$ be a nonempty subset of $A'$. Set
$$i + B = \{i + a_1 \pmod{p}, i + a_2 \pmod{p}, \ldots, i + a_k \pmod{p}\}$$
Note that the sums of the elements of the sets $i + B$, $i = 0, 1, \ldots, p-1$, are all distinct. Indeed, if for some $s$ and $t$ the sums are equal then
$$\begin{aligned} & \sum_{i=1}^{k} (s + a_i) \equiv \sum_{i=1}^{k} (t + a_i) \pmod{p} \\ \Longleftrightarrow \ & k s + \sum_{i=1}^{k} a_i \equiv k t + \sum_{i=1}^{k} a_i \pmod{p} \\ \Longleftrightarrow \ & k s \equiv k t \pmod{p} \end{aligned}$$
which is equivalent to $s = t$.

Therefore the set of the subsets of $A'$ (without the empty set and $A'$) partitions into $\frac{2^p - 2}{p}$ groups and every group contains $p$ sets. Moreover, the sums of the elements of the subsets in every group run over all residues modulo $p$.

Therefore the number of the subsets having sums divisible by $p$ equals $\frac{2^p - 2}{p}$. Since $0$ is included in half of them, it follows that the number of the subsets $B$ of $A$ (including the empty set and excluding $A$) equals $\frac{2^p - 2}{2p}$.

Replacing the empty set by $A$ (having sum $\frac{p(p-1)}{2}$, which is divisible by $p$), we conclude that the answer is $\frac{2^{p-1} - 1}{p}$.