Show that if $SAT \in P/klog(n)$ then $SAT \in P$
You need to combine two ideas here:

There are only polynomially many possible advice strings.
If you can solve the decision version of SAT, then you can solve the search version of SAT.

There is a slight annoyance that you need the search-to-decision reduction to involve only constantly many different input lengths. Assuming for simplicity that you can perform everything using a single input length $n$, here is what you do: you go over all possible advices for length $n$, and for each one, either determine that the SAT instance is unsatisfiable, or attempt to find a satisfying assignment. If you find a satisfying assignment, then the SAT instance is satisfiable, otherwise it isn't.
I'll let you fill in the details.