So, there's a deputy. called $P$. who is a member of at least $\Big\lceil \frac{450}{100}\Big\rceil =5$ commissions. named $C_1,C_2,C_3,C_4,C_5$. Suppose to the contrary that there not exist $4$ commissions which exactly one common deputy. Consider any four-element distinct subsets $S_1,S_2$ of $\{ 1,2,3,4,5\}$, there exist deputies $P_1$ and $P_2$, both is not the same with $P$, who, respectively, is other common deputy of $C_{s_1}$ where $s_1\in S_1$ and $C_{s_2}$ where $s_2\in S_2$. If $P_1=P_2$, we get that $P_1\in C_s$ for all $s\in S$ since $S=S_1\cup S_2$, which means $P_1$ is other common deputy of $C_s$ where $s\in S$, contradiction. Hence $P_1\neq P_2$. Let $Q_1,Q_2,Q_3$ be the other common deputy of the sets $\{ C_1,C_2,C_3,C_4\} ,\{ C_1,C_2,C_3,C_5\} ,\{ C_1,C_2.C_4,C_5\}$. Note that $Q_1,Q_2,Q_3,P$ are pairwise distinct, and all of them are common deputies of $C_1,C_2$, contradiction.