One way to get the SoP form starts by multiplying everything out, using the distributive law:
$$\begin{align*}
(ac+b)(a+b'c)+ac&=ac(a+b'c)+b(a+b'c)+ac\\
&=aca+acb'c+ba+bb'c+ac\\
&=ac+ab'c+ab+ac\\
&=ac+ab'c+ab\;.
\end{align*}$$
Then make sure that every term contains each of $a,b$, and $c$ by using the fact that $x+x'=1$:
$$\begin{align*}
ac+ab'c+ab&=ac(b+b')+ab'c+ab(c+c')\\
&=abc+ab'c+ab'c+abc+abc'\\
&=abc+ab'c+abc'\;.
\end{align*}$$
Alternatively, you can make what amounts to a truth table for the expression:
$$\begin{array}{cc}
a&b&c&ac+b&b'c&a+b'c&ac&(ac+b)(a+b'c)+ac\\ \hline
0&0&0&0&0&0&0&0\\
0&0&1&0&1&1&0&0\\
0&1&0&1&0&0&0&0\\
0&1&1&1&0&0&0&0\\
1&0&0&0&0&1&0&0\\
1&0&1&1&0&1&1&1\\
1&1&0&1&1&1&0&1\\
1&1&1&1&0&1&1&1
\end{array}$$
Now find the rows in which the expression evaluates to $1$; here it’s the last three rows. For a product for each of those rows; if $x$ is one of the variables, use $x$ if it appears with a $1$ in that row, and use $x'$ if it appears with a $0$. Thus, the last three rows yield (in order from top to bottom) the terms $ab'c$, $abc'$ and $abc$.
You can use the truth table to get the PoS as well. This time you’ll use the rows in which the expression evaluates to $0$ — in this case the first five rows. Each row will give you a factor $x+y+z$, where $x$ is either $a$ or $a'$, $y$ is either $b$ or $b'$, and $z$ is either $c$ or $c'$. This time we use the variable if it appears in that row with a $0$, and we use its negation if it appears with a $1$. Thus, the first row produces the sum $a+b+c$, the second produces the sum $a+b+c'$, and altogether we get
$$(a+b+c)(a+b+c')(a+b'+c)(a+b'+c')(a'+b+c)\;.\tag{1}$$
An equivalent procedure that does not use the truth table is to begin by using De Morgan’s laws to negate (invert) the original expression:
$$\begin{align*}
\Big((ac+b)(a+b'c)+ac\Big)'&=\Big((ac+b)(a+b'c)\Big)'(ac)'\\
&=\Big((ac+b)'+(a+b'c)'\Big)(a'+c')\\
&=\Big((ac)'b'+a'(b'c)'\Big)(a'+c')\\
&=\Big((a'+c')b'+a'(b+c')\Big)(a'+c')\\
&=(a'b'+b'c'+a'b+a'c')(a'+c')\\
&=a'b'(a'+c')+b'c'(a'+c')+a'b(a'+c')+a'c'(a'+c')\\
&=a'b'+a'b'c'+a'b'c'+b'c'+a'b+a'bc'+a'c'+a'c'\\
&=a'b'+a'b'c'+b'c'+a'b+a'bc'+a'c+a'c'\\
&=a'b'+b'c'+a'b+a'(c+c')\\
&=a'b+b'c'+a'b+a'\\
&=b'c'+a'\;,
\end{align*}$$
where in the last few steps I used the absorption law $x+xy=x$ a few times. Now find the SoP form of this:
$$\begin{align*}
b'c'+a'&=b'c'(a+a')+a'(b+b')(c+c')\\
&=ab'c'+a'b'c'+a'b(c+c')+a'b'(c+c')\\
&=ab'c'+a'b'c'+a'bc+a'bc'+a'b'c+a'b'c'\\
&=ab'c'+a'b'c'+a'bc+a'bc'+a'b'c\;.
\end{align*}$$
Now negate (invert) this last expression, and you’ll have the PoS form of the original expression:
$$\begin{align*}
(ab'c'&+a'b'c'+a'bc+a'bc'+a'b'c)'\\
&=(ab'c')'(a'b'c')'(a'bc)'(a'bc')'(a'b'c)'\\
&=(a'+b+c)(a+b+c)(a+b'+c')(a+b'+c)(a+b+c')\;,
\end{align*}$$
which is of course the same as $(1)$, though the factors appear in a different order.
It might help to think by analogy with algebraic expressions formed using multiplication and addition. Multiplication distributes through addition, i.e., you have $x \times (y + z) = (x \times y) + (x \times z)$. Hence you can write any expression as a sum of products of atoms, by pushing the multiplications in through additions. But note that addition does not distribute through multiplication, so you can't write every expression as a product of sums of atoms.
Boolean algebra, however is much more symmetric: not only does conjunction distribute through disjunction, i.e., $x \land (y \lor z) = (x \land y) \lor (x \land z)$ but also disjunction distributes through conjunction, i.e., $ x \lor (y \land z) = (x \lor y) \land (x \lor z)$. Hence (also using De Morgan's laws to push in negations), you get to choose whether to push the conjunctions in, which will lead you to DNF: a disjunction of conjunctions of literals, or to push the disjunctions in, which will lead you to a CNF: a conjunction of disjunctions of literals. (Here "literal" means an atom or negated atom.) For example, your formula can be transformed into a CNF like this:
$$
\begin{array}{rcl}
[(p \land q) \lor \lnot p] \lor \lnot q &=& [(p \lor \lnot p) \land (q \lor \lnot p)] \lor \lnot q\\
&=& (p \lor \lnot p \lor \lnot q) \land (q \lor \lnot p \lor \lnot q)
\end{array}
$$
You can now do further simplifications if you wish, e.g., to arrive at $p \lor \lnot p$ (which is both a DNF and a CNF).
Best Answer
Yes, that's right:
(B+C')(B+A')(C'+A) = (Double negation)
[(B+C')(B+A')(C'+A)]'' = (DeMorgan)
[(B+C')'+(B+A')'+(C'+A)']'
... which is exactly your
$$ \overline{\overline{(B+C')}+\overline{(B+A')}+\overline{(C'+A)}} $$
Also, if you can't work with C', and have to work with C (that is, if you can't use Not's), use the fact that:
C' = C NOR C
... Unfortunately what you did at the beginning is not right ...
(A + B)(B'C)(A' + C') does not work out to
(B'C)+(B'A)+(CA')
and you certainly can't just negate that to do a DeMorgan on that to get:
(B+C')(B+A')(C'+A)
since you'd just be adding a negation to the whole expression out of nowhere!
Instead:
(A + B)(B'C)(A' + C') = (Association)
(A + B)B'C(A' + C') = Reduction x 2
AB'CA' = False!
And an easy NOR formula for that is: NOR(A,A') = NOR(A,NOR(A,A))
I wonder though ... should that initial expression maybe be:
F = (A + B)(B'+C)(A' + C') ?
Because then you can do what I did above:
(A + B)(B'+C)(A' + C') =
[(A + B)(B'+C)(A' + C')]'' =
[(A + B)'+(B'+C)'+(A' + C')']' =
NOR(NOR(A,B'),NOR(B',C),NOR(A',C')) = (if needed)
NOR(NOR(A,NOR(B,B)),NOR(NOR(B,B),C),NOR(NOR(A,A),NOR(C,C)))