Clearly if $A \subseteq K$ and $A \in F$ then $K \setminus A \not \in F$. Therefore from any pair of the collection of pairs $\{A, K \setminus A\}$, with $A$ a subset of $K$, exactly one element belongs to $F$, since both collections have cardinality $16$. From the pair $\{\emptyset, K\}$ clearly $K \in F$. Consider the pairs where $A$ is a singleton; if one such is in $F$, then it is the only one, and all sets of $F$ contain the singleton of $A$. Assume therefore all complements of singletons are in $F$. Consider now the pairs where $A$ is a doubleton; if two such are in $F$, then their intersection must be a singleton, say $\{a\}$, but then the further intersection with $\{b,c,d,e\}$ from $F$ is empty - contradiction. So at most one doubleton is in $F$, say $\{a,b\}$. Then all other tripletons but $\{c,d,e\}$ are in $F$, among them $\{a,c,d\}$ and $\{b,c,e\}$, with intersection $\{c\}$, but then the further intersection with $\{a,b,d,e\}$ from $F$ is empty - contradiction.