 Solution

Distinct traces give the bound. If the common-set alternative fails, the preceding argument shows that the traces $B_j$ are all distinct. They form a set family on the $k$-element ground set $K$ with constant pairwise intersection $\lambda >0$. The constant-intersection family bound therefore gives $q\le k$. Since $q=m-t+2$,Here $k\ge \lambda >0$, because any two of the remaining traces intersect in $\lambda$ elements, so applying the bound to $K$ is legitimate.

The bound is sharp even when the common-set alternative fails. For any $m\ge t+1$, take ground set $[m]$ and $A_i=[m]\setminus \{i\}$ for $1\le i\le m$. Every $t$-fold intersection has size $\lambda =m-t>0$, but the intersection of all members is empty. Every $(t-2)$-fold intersection has size $k=m-t+2$. Thusand there is no common subset of size $\lambda$. These complement-of-singleton extremizers show that the bound cannot be reduced in general.