How to simplify the next boolean algebra expresion?
Some details are off in the application of De Morgans' laws.

De Morgans' laws state that

$$ \overline{(A + B)} = (\overline{A} * \overline{B}) $$
and
$$ \overline{(A * B)} = (\overline{A} + \overline{B})$$
(using the multiplication symbol * to stress what is being negated)

So in your case first of all

\$B(\overline{A}C + A\overline{C}) \neq B(\overline{A\overline{C}} + \overline{\overline{A}C})\$

But instead

\$\overline{A}C = \overline{A} * C = \overline{A} * \overline{\overline{C}} = \overline{(A + \overline{C})}\$

And

\$A\overline{C} = A * \overline{C} = \overline{\overline{A}} * \overline{C} = \overline{(\overline{A} + C)}\$

So

\$B(\overline{A}C + A\overline{C}) = B(\overline{(A + \overline{C})}+\overline{(\overline{A} + C)})\$

And applying De Morgan to the sum in the outer parentheses we get

\$B(\overline{(A + \overline{C})}+\overline{(\overline{A} + C)})=B(\overline{(A + \overline{C}) *  (\overline{A} + C)}) \$

That is

\$B(\overline{A\overline{A} +AC + \overline{C}\overline{A} + \overline{C}C}) = B(\overline{AC + \overline{C}\overline{A}})\$

Things start smelling a bit, looks like we might get back at the initial formula, let's see where we get from here

\$B(\overline{AC + \overline{C}\overline{A}}) = B(\overline{AC} * \overline{\overline{C}\overline{A}})\$

And similarly to what we did before, we get

\$\overline{AC} = (\overline{A} + \overline{C})\$

And

\$\overline{\overline{C}\overline{A}} = (C + A)\$

So

\$B(\overline{AC} * \overline{\overline{C}\overline{A}}) = B((\overline{A} + \overline{C}) * (A + C)) = B(\overline{A}A + \overline{A}C + \overline{C}A + \overline{C}C) = B(\overline{A}C + \overline{C}A)\$

That's it, we're back to square one.

As other have answered before you can see \$(\overline{A}C + A\overline{C})\$ as the exclusive or between \$A\$ and \$C\$, but you won't have it any simpler than that.