 Solution

A repeated trace produces the common set. If $B_j=B_{\ell }$ for two distinct remaining indices, their common value $S$ has size $\lambda$. For any other remaining index $h$, the identity $|B_h\cap B_j|=\lambda =|S|$ forces $S\subseteq B_h$. The inclusion also holds for $j,\ell$ themselves. Since $S\subseteq K$, it lies in every $A_i$ indexed by $I$ as well. ConsequentlyThis proves the common-set alternative of the constant t-wise intersection dichotomy. The conclusion only requires a common subset of size $\lambda$; for $m\ge t$, the full common intersection in fact also has size $\lambda$, because it is contained in a $t$-fold intersection of that size.