It's usually easier to figure these things out with Karnaugh maps. After playing around, we see that:
\begin{align*}
\overline B + \overline C(B + A)
&= \overline B(1) + \overline C(B + A) \\
&= \overline B(1 + \overline C) + \overline C(B + A) \\
&= \overline B + \overline B ~ \overline C + \overline C(B + A) \\
&= \overline B + \overline C(\overline B + B + A) \\
&= \overline B + \overline C(1 + A) \\
&= \overline B + \overline C(1) \\
&= \overline B + \overline C
\end{align*}
as desired.
HINT
This equivalence principle will be your friend:
Adjacency
$PQ + PQ' = P$
If you're not allowed to use Adjacency in $1$ step, here is a derivation of Adjacency in terms of more basic equivalence principles:
$$P Q + (P Q') \overset{Distribution}=$$
$$P (Q + Q') \overset{Complement}=$$
$$P 1 \overset{Identity}=$$
$$P$$
To apply Adjacency, note that $P$ and $Q$ can be any complex expressions, so in this case, where every terms has $4$ variables, just look for two terms that are the same for $3$ of the variables, but differ in the fourth. For example, the first two terms are the same except for the $D$ variable, so those can be combined:
$A'BC'D'+A'BC'D=A'BC'$
You can also combine the first and seventh terms:
$A'BC'D'+ABC'D'=BC'D'$
To do both of those, you would need to 'reuse' the first term, but you can get as many copies as you want by:
Idempotence
$P + P = P$
So, for example, focusing on the first, second, and seventh term:
$$A'BC'D'+A'BC'D+ABC'D'\overset{Idempotence}=$$
$$A'BC'D'+A'BC'D'+A'BC'D+ABC'D'\overset{Commutation}=$$
$$A'BC'D'+A'BC'D+A'BC'D'+ABC'D'\overset{Adjacency \ x \ 2}=$$
$$A'BC'+BC'D'$$
Best Answer
Notice that $$A\bar C=A\bar C(B+\bar B)=A\bar CB+A\bar C\bar B$$ Therefore $$\begin{align}A\bar C+A\bar B+\bar CB&=A\bar CB+A\bar C\bar B+A\bar B+\bar CB\\&=A\bar B(\bar C+1)+\bar CB(A+1)\\&=A\bar B+\bar CB\end{align}$$