Plat representations of unlinks Suppose that $\beta$ is a $2n$-strand braid with plat closure $L$.  We can multiply $\beta$ on either side by a member of the Hilden subgroup to get a new braid whose plat closure is still $L$.  Or we could replace $\beta$ with $\beta\sigma_{2n-1} \in B_{2n+2}$ to get another braid whose plat closure is $L$.  (Here $\sigma_{2n-1}$ is one of the usual Artin generators.)
In "On the Stable Equivalence of Plat Representations of Knots and Links", Birman proves that these moves are 'sufficient'.

Theorem: Let $K$ and $K'$ be knots.  Let $\alpha$ and $\beta$ be braids whose plat closures are isotopic to $K$ and $K'$ respectively. Then $K$ is isotopic to $K'$ if and only if there are stabilizations $\alpha'$ and $\beta'$ of $\alpha$ and $\beta$ which lie in the same Hilden double coset.

(The theorem is still true for links with one stabilization operation for each component of the closure.)
Montesinos exhibited two minimal index braids whose plat closures are isotopic but which do not lie in the same Hilden double coset.  So stabilization is necessary.
Now suppose that $\beta$ and $\beta'$ are plat representations of the $k$-component unlink.  Is anything known about whether stabilization is necessary?
 A: Otal showed that stabilization is not needed for the unknot: any bridge representative of the unknot is equivalent to a stabilization
of the trivial 1-bridge representative. This theorem can be regarded a variant of Waldhausen's theorem that every Heegaard splitting of the 3-sphere is a stabilization of the trivial one. The connection is to note that a cyclic branched cover of the unknot is the 3-sphere again, with preimage of the bridge surface a Heegaard splitting. This Heegaard splitting is not only a stablization of the trivial one, but Otal's theorem implies that it is an equivariant stabilization with respect to the finite cyclic action. 
For the unlink, I found a paper which shows in Corollary 3.4 that any $n$-bridge surface of an unlink has an $n$-bridge representative which is obviously split (the components are separated by spheres intersecting the bridge sphere in a single loop). The statement is a bit ambiguous though as to whether the bridge spheres are equivalent (although I interpreted this to be what they meant, because the alternative is trivial: any unlink on $n$ components has a bridge representation with this property). In any case, the result is not surprising, since it generalizes (by the branched cover trick) Haken's lemma, which states that a Heegaard splitting of a reducible 3-manifold is reducible (has a 2-sphere intersect it in a single point). In fact, I think the same proof ought to work for plats as for Heegaard splittings, although I haven't checked it carefully. 
