Option $(D)$ is correct here.
- $A.\bar A=1$ is false. $A.\bar A=0$ is correct.
- $A+AB=A+B$ is also false. $A(1+B)\implies A.1=A$
- $A(A+B)=B$ is false. $(A+AB)$ can be written as option $2$
- $A+\bar AB=A+B$ is true using distributive law.
$A+\bar AB\implies (A+\bar A)(A+B)\implies (A+B)$