Suppose $(\Sigma,\phi)$ is an abstract open book whose monodromy is expressed as a product of Dehn twists about boundary-parallel curves. Is there a standard way to produce a surgery presentation of the resulting 3-manifold?
(I have such a procedure in mind, but I haven't checked through the details rigorously. For the case where $\phi=\operatorname{id}$, it's roughly as follows: Let $\hat \Sigma$ be the closed surface obtained by capping off the $n$ boundary components of $\Sigma$ with disks $D_i$, where $1 \leq i \leq n$. The mapping torus of the identity map on $\hat \Sigma$ is just $\hat \Sigma \times S^1$. Starting with $\hat \Sigma \times S^1$, we obtain the manifold given by the open book $(\Sigma,\operatorname{id})$ by drilling out the solid tori $D_i \times S^1$ and gluing in copies of $S^1 \times D^2$ (note the reversal of factors). This is simply $0$-surgery on the core of each $D_i \times S^1$. Now start with the standard Kirby diagram for $\hat \Sigma \times D^2$, whose boundary is $\hat \Sigma \times S^1$. There is a unique 2-handle in the diagram, and the aforementioned Dehn surgeries correspond to adding $0$-framed 2-handles along meridians of the original 2-handle's attaching circle. The boundary now has an open book decomposition $(\Sigma,\operatorname{id})$. If we want a genuine surgery description on a link in $S^3$, we can take the Kirby diagram and turn the 1-handles into dotted circles and then 0-framed unknots. In the case where the monodromy is expressed as a product of Dehn twists about boundary-parallel curves, I believe we take the $0$-framed meridians in the Kirby diagram and change their framing to reflect the number of Dehn twists about the corresponding boundary component.)