= Solution
\b[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 $q\leq k$. Since $q=m-t+2$,
$$
\boxed{m\leq k+t-2.}
$$
Here $k\geq\lambda>0$, because any two of the remaining traces intersect in $\lambda$ elements, so applying the bound to $K$ is legitimate.
\b[The bound is sharp even when the common-set alternative fails.] For any $m\geq t+1$, take ground set $[m]$ and $A_i=[m]\setminus\{i\}$ for $1\leq i\leq m$. 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 $k=m-t+2$. Thus
$$
\boxed{m=k+t-2,}
$$
and there is no common subset of size $\lambda$. These <complement-of-singleton extremizers> show that the bound cannot be reduced in general.
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