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Use the definitions in L29. Each of the $I(P)$ triples extends in $n-3$ ways to a bad four-set. Each four-set contains at most four such triples. Therefore

$$(n-3)I(P)\le4D(P).$$

In every bad four-set, choose two distinct unordered pairs of equal length. If they share an endpoint, their three endpoints are one of the triples counted by $I(P)$. All four-sets containing any such triple number at most $(n-3)I(P)$. Otherwise their four endpoints are distinct, and the equality is counted by $Q(P)$. Each ordered quadruple has only one underlying four-set, so the number of four-sets of this second type is at most $Q(P)$. Overlap between the types only enlarges this upper bound.