I do not know if there is a "standard/correct/natural" method of defining the notion of dual bimodule. A little search i made in the literature did not lead me to smt "standard". In any case, expanding the idea in my comment above, here is what i have thought of:
Given an $A$-$B$-bimodule $M$ we can always view this as a right $B\otimes A^{op}$-module setting $m(b\otimes a^{op})=amb$. Conversely, if $M$ is a right $B\otimes A^{op}$-module then we get an $A$-$B$-bimodule by setting $an=m(1\otimes a^{op})$ and $nb=n(b\otimes 1)$.
(Here $a^{op}$ stands for the element $a$ of the algebra $A$ viewed as an element of the opposite algebra/ring $A^{op}$).
So, start with your $A$-$B$-bimodule, view it as a right $B\otimes A^{op}$-module and then get the dual module $Hom_{B\otimes A^{op}}(M,B\otimes A^{op})$. This will be a left $B\otimes A^{op}$-module. Following a strategy similar to the one mentioned in the preceding paragraph this can be viewed as an $A^{op}$-$B^{op}$-bimodule, which may be what you are looking for.
This is a general method, which i think works for modules over rings or algebras and it does not have something to do with the projective case.