Why is $A' + AB' + B$ always true?

You have $$AB^\prime = (AB^\prime)^{\prime \prime}=(A^\prime+B^{\prime \prime})^\prime = (A^\prime + B)^\prime$$ by double negation and de Morgan law and therefore

$$A^\prime+AB^\prime+B=(A^\prime+B) + AB^\prime = (A^\prime+B)+(A^\prime + B)^\prime = 1$$

A hint: by de Morgan,

$$A'+AB'+B = (A(A'+B)B')'$$

Now see if the expression inside the outer parentheses simplifies.

\begin{align} A'+AB'+B \tag{1}\label{1} \end{align}

Since

\begin{align} A+A'=B+B'&=1 \tag{2}\label{2} , \end{align}

we can rewrite \eqref{1} as

\begin{align} &A'(B+B')+AB'+B(A+A') \\ &=A'B+A'B'+AB'+AB \tag{3}\label{3} \\ &=A'(B+B')+A(B'+B) \\ &=A'+A=1 \tag{4}\label{4} . \end{align}