How can a Kirby diagram fail to determine a handle decomposition? I've read that a handle decomposition for 4-manifold determines a unique smooth structure, and I've also read that every 4-manifold admits a Kirby diagram.  So when does a Kirby diagram fail to determine a handle decomposition?
This question occurred to me while looking at the standard Kirby diagram for the $E_8$-manifold.
 A: It is not true that every topological $4$-manifold admits a Kirby diagram. In fact, a handle-decomposition of a $4$-manifold and a Kirby diagram for that $4$-manifold are equivalent notions. The Kirby diagram is exactly the attaching data for the handles. By a theorem of Morse, every smooth $n$-manifold admits a handle-decomposition so in particular every smooth $4$-manifold admits a Kirby diagram. Locally-flat embeddings can always be smoothed in dimensions $ \leq 3$, so as the attaching is being done in dimension $3$ one can always isotope the attaching maps to be smooth for each attached handle. In particular a topological handle-decomposition on a $4$-manifold can be refined to a smooth one, and hence a topological $4$-manifold with a handle-decomposition must admit a smooth structure coming from this decomposition. (As far as I know the source for this argument is due to Kirby. It appears here.) 
However, it is not true that every topological $4$-manifold admits a smooth structure (i.e. Freedman's E8 manifold given by capping the E8 plumbing by a contractible 4-manifold in Mike Miller's comment). In particular, such a manifold cannot admit a topological handle-decomposition (or a Kirby diagram).
