Complement splitting for forbidden midpoint intersections ($|\mathcal{A}|\le 2\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$)

For a prime $p$, take rank-$2p$ subsets of $[4p]$ with no distinct pair meeting in $p$ points. Split by membership of coordinate one. In the containing part, remove that coordinate; in the avoiding part, first take complements and then remove it. Both transformed families have rank $2p-1$ on $4p-1$ points. Their distinct intersections avoid residue $p-1$ modulo $p$. The uniform Frankl-Wilson theorem bounds each by $\left(\genfrac{}{}{0ex}{}{4p-1}{p-1}\right)$, yielding the displayed stronger form of the requested bound.

Fix a $(p+1)$-set $B\subseteq [4p]$. Take every $2p$-set $B\cup X$ with $X$ a $(p-1)$-subset of the remaining $3p-1$ points, together with the complements of all these sets. Within either half, intersections have size at least $p+1$. Across the halves they have size $|X\setminus Y|\le p-1$. Thus no intersection has size $p$, and the two disjoint halves give the displayed family size.