Just as when proving trig identities, in general it’s better to start by trying to simplify the more complicated side, which in this case is the right-hand side. If you expand the RHS by brute force and use the identities $xx=x$, $x\bar{x}=0$, $x+\bar{x}=1$, and $x+xy=x$ repeatedly, you shouldn’t have much trouble reducing it to $$x_1x_2+x_1\bar{x}_3+x_2x_3+\bar{x}_2\bar{x}_3,$$ which is almost what you want. Now write $$x_1x_2 = x_1x_2 1 = x_1x_2 (x_3 + \bar{x}_3),$$ expand, and use the absorption identity again to get rid of the extra terms.
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.
Best Answer
Yes: multiplication distributes over products, so $$(a'+b+d')(a'+b+c'+f') = a'(a'+b+c'+f') + b(a'+b+c'+f') + d'(a'+b+c'+f'),$$ and you can then distribute again each of the factors on the right.
You should then simplify in any number of ways; for example, you have $a'b$ and $ba'$, and since $a'b+a'b = a'b$, you can drop one of them. Since $bb=b$, you can rewrite $bb$ as $b$; etc.
In the end, you should have a sum of products. Then you can just invert.
For example, if at the end you had $$ p' = a'b + bc' + d'f' + a'f'$$ (you won't, but say you did), then you would have $$p = p'' = (a'b + bc' + d'f' + a'f')' = (a'b)'\cdot(bc')'\cdot (d'f')'\cdot (a'f')'$$ and then you would apply De Morgan's laws to each of the factors, e.g., $(a'b)' = a+b'$, so you would have $$p = (a+b') \cdot (b'+c)\cdot (d+f)\cdot (a+f),$$ a product of sums.