Just to check: Since your question is flagged VASP, I assume you are simulating under periodic boundary conditions, but since you write of a slab, I assume that your simulation cell has a form "VABV" where V stands for some sufficiently large vacuum. (Otherwise, BAB and AB "slabs" without vacuum would be equivalent except for the thickness of the B-regime, and result in the same interfaces.)

Assuming that between the simulations the volume of individual A and B slabs is equivalent, by simulating the joined slabs you're removing one A-surface and one B-surface, while adding one AB interface. So your formula really would give you 

• Einterface(AB) - Esurface(A) - Esurface(B)

assuming that by "1/A" you mean the "1/area on one side of the slab", which is probably the same as "1/cross-section of the simulation cell". 

Therefore I'd say that you need the individual surface energies, which requires comparing the energy per atom in bulk configuration to your slab simulations. 

At that point however, you can just calculate the interface energy from the AB slab and I would see now reason to simulate an ABA slab at all. Both orthorhombic and fcc crystals should have inversion symmetry, so the interfaces and surfaces in your slabs should be equivalent ( Eif(AB) = Eif(BA), Esf(VA) = Esf(AV), etc.).

Unless of course the BA and the AB interface are not non-equivalent due to e.g. different orientation of the BAB slabs, but even then it is likely easier and yields better accuracy for the same effort to calculate the interface energies separately from a BA and an AB simulation. The interface energy is probably even easier to calculate from a gain-boundary simulation (i.e. AB without vacuum), though you may have to think about relaxing the length of the simulation cell normal to the interface (which may be easier with a slab). 

In our publication that resulted from my diploma thesis, we called the relevant properties γGB and γsurf. Since it is some time past and I haven't worked with interface simulations since then, I'm not sure though if there are better references for defining the grain-boundary energy. 

As a general approach: Think about how you can model the total energy of the system as a linear superposition of bulk, surface, and interface energies, e.g. 

• Etot(AB) = Ac Esf(A) + NA Ebk(A) + Ac Eif(AB) + NA Ebk(B) + Ac Esf(B)

where Ac is the cross-section of the simulated cell and NA the number of atoms in each slab.

and then think about finding combinations of simulations that eliminate all but one of the terms. This almost certainly means starting with bulk simulations, which give Ebk pretty directly (Etot(bulk) = Nbulk Ebk). 

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Hi Klaus-Dieter Bauer

Thank you very much for your kind reply.  You are right I should have write my two structure as ABV (where v is the vacuum) and BABV.

Now I am still confuse about the surface energy of both?

I know for ABV

The formula of Interface Energy = (1/A){E(AB)slab-E(Aslab)-E(Bslab)}. holde good

Then the formula for BABV should be ?

I updated my answer a little. Forgot to post the edits before leaving for lunch. It contains now a short explanation how to get such formulas in the first place. 

Regarding your formula: Except for the sign it is the energy needed to split the interface, resulting in two additional surfaces, which differs from what is usually called the interface energy.