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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 $am=m(1\otimes a^{op})$ and $mb=m(b\otimes 1)$.
(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.

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 $am=m(1\otimes a^{op})$ and $mb=m(b\otimes 1)$.
(Here $a^{op}$ stands for the element $a$ of the algebra $A$ viewed as an element of the opposite algebra/ring $A^{op}$).
Similarly, an $A$-$B$-bimodule $M$ can be viewed as a left $B^{op}\otimes A$ module.

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, or equivalently a $B$-$A$-bimodule which may be the notion of the dual of your initial $A$-$B$-bimodule you are looking for.

P.S.: This is a general method, which i think works for modules over rings or algebras. It is not tied especially to the projective case.

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.

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 $am=m(1\otimes a^{op})$ and $mb=m(b\otimes 1)$.
(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.

I do not know if the following is a "standard/correct/natural" method of defining the dual bimodule but here is what i have thought:

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.
(I will try to add more details asap).

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.