Subsets whose sum is divisible by p

Roots of unity filter. Let $N$ be the number we want. For a subset $A$ with sum $\Sigma(A)$,
$$\frac1p\sum_{j=0}^{p-1}\omega^{j\Sigma(A)}=\begin{cases}1&p\mid\Sigma(A)\\0&\text{otherwise,}\end{cases}$$
so
$$N=\frac1p\sum_{j=0}^{p-1}\sum_{A}\omega^{j\Sigma(A)}=\frac1p\sum_{j=0}^{p-1}\prod_{k=1}^p\bigl(1+\omega^{jk}\bigr).$$

The term $j=0$ gives $2^p$.

The terms $j=1,\dots,p-1$. The residues $jk$ modulo $p$ run over all residues exactly once, so
$$\prod_{k=1}^p(1+\omega^{jk})=\prod_{r=0}^{p-1}(1+\omega^r).$$
From $\prod_{r=0}^{p-1}(x-\omega^r)=x^p-1$ at $x=-1$:
$$\prod_{r=0}^{p-1}(1+\omega^r)=2.$$

Conclusion.
$$N=\frac1p\bigl(2^p+(p-1)\cdot2\bigr)=\frac{2^p+2(p-1)}p.$$

Check. $p=3$: $\frac{8+4}3=4$. $p=5$: $8$. $p=7$: $20$.

Remark. As a corollary, $2^p\equiv2\pmod p$, a proof of Fermat's little theorem for base $2$.

Lesson. To count objects with a divisibility condition, average a root of unity over the residues.