Electric potential obeys superposition principle so potential due to electric dipole as a whole would be sum of potential due to both the charges +q and -q. Thus

where r₁ and r₂ respectively are distance of charge +q and -q from point R.

Now draw line PC perpandicular to RO and line QD perpandicular to RO as shown in figure. From triangle POC
cosθ=OC/OP = OC/a
therefore OC=acosθ similarly OD=acosθ
Now ,
r₁ = QR≅RD = OR-OD = r-acosθ
r₂ = PR≅RC = OR+OC = r+acosθ

since magnitude of dipole is 
|p| = 2qa

If we consider the case where r>>a then

again since pcosθ= p·rˆ where, rˆ is the unit vector along the vector OR then electric potential of dipole is