According to Boolean algebra, which one of the following is a correct statement.
A) ${\left( {a + b} \right)^\prime } = a' + b'$
B) ${\left( {a + b} \right)^\prime } = a' \cdot b'$
C) ${\left( {a \cdot b} \right)^\prime } = \left( {a' \cdot b'} \right)$
D) None of these

- Hint: This question is based on the inversion laws on addition and multiplication, which are stated by the de Morgan’s theorem. So, apply de Morgan’s theorem on each and every option’s LHS(Left Hand Side) then compare the value derived by its RHS(Right Hand Side), if it satisfies then we’ll get our answer.

Complete step-by-step solution:
Here first of all we should know that according to the de Morgan’s theorem of Boolean algebra,
$ {\left( {a + b} \right)^\prime } = a' \cdot b' \\
  {\left( {a \cdot b} \right)^\prime } = a' + b' \\ $ }…….de Morgan’s theorem
So, now first of all we will check option A by comparing their Right hand side (RHS) and Left hand side (LHS)
LHS$ = {\left( {a + b} \right)^\prime } = a' \cdot b'$ …(According to de Morgan’s theorem)
  As we can see that,
LHS $ \ne $ RHS
i.e $a' \cdot b' \ne a' + b'$
$ \Rightarrow $Option A is incorrect.
Now similarly we will check option B by comparing their Right hand side (RHS) and Left hand side (RHS)
LHS $ = {\left( {a + b} \right)^\prime } = a' \cdot b'$ …(According to de Morgan’s theorem)
Here we can see that,
LHS$ = $RHS
i.e $a' \cdot b' = a' \cdot b'$
$ \Rightarrow $Option B is correct.
Now again we will repeat the procedure for option C,
LHS $ = {\left( {a \cdot b} \right)^\prime } = \left( {a' + b'} \right)$ ….(According to de Morgan’s theorem)
Here we can see that,
LHS $ \ne $ RHS
i.e $a' + b' \ne \left( {a' \cdot b'} \right)$
$ \Rightarrow $Option C is incorrect.

Therefore, the correct option is option B.

Note:
For solving this question faster and easily like in competitive exams where time is limited you can check the option by putting the values of variables $a$ and $b$ as $0$ and $1$ or $1$ and $0$ and then you have to compare Left hand side with the Right hand side.