= Solution
We use the coefficient form of the <Combinatorial Nullstellensatz>: if a polynomial of degree $\sum_i t_i$ has nonzero coefficient at $\prod_i x_i^{t_i}$, then it cannot vanish on every product set $\prod_iS_i$ with $|S_i|>t_i$.
Write the terms as $(a_i,b_i)$ and, over $\mathbb F_p$, put
$$
A=\sum_i a_ix_i,
\qquad B=\sum_i b_ix_i,
\qquad S=\sum_i x_i,
$$
and
$$
E_p(x)=\sum_{|J|=p}\prod_{j\in J}x_j.
$$
Consider
$$
F(x)=
(1-A^{p-1})(1-B^{p-1})
\left[(1-S^{p-1})(2-E_p(x))
-2\prod_i(1-x_i)\right].
$$
A direct multinomial-coefficient calculation, using <Wilson theorem>[Wilson's theorem], shows that the coefficient of $\prod_{i=1}^{4p-2}x_i$ in $F$ is nonzero. The Combinatorial Nullstellensatz with every $S_i=\{0,1\}$ therefore gives an indicator vector $x$ for which $F(x)\ne0$.
The zero indicator does not work. The first two factors force the selected vectors to have both coordinate sums zero. If the support size is not divisible by $p$, <Fermat's little theorem> makes $1-S^{p-1}=0$, while the final product also vanishes; hence the support has size $p$, $2p$, or $3p$. Size $p$ proves the claim, and size $3p$ reduces to part (a). For size $2p$,
$$
E_p(x)=\binom{2p}{p}\equiv2\pmod p,
$$
so the bracket vanishes, again contradicting $F(x)\ne0$. Only the size-$p$ and reducible size-$3p$ cases remain, and either yields the required zero-sum $p$-subsequence.
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