(B) The statement is True.
Step $1$: Evaluate the intersection $B \cap \varnothing$. Since the intersection of any set with the empty set is the empty set,$B \cap \varnothing = \varnothing$.
Step $2$: Substitute this into the expression: $A \times \{B \cap \varnothing\} = A \times \{\varnothing\}$.
Step $3$: The Cartesian product $A \times \{\varnothing\}$ is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in \{\varnothing\}$. This results in $\{(1, \varnothing), (2, \varnothing)\}$.
Wait,let us re-evaluate the original expression: $A \times \{B \cap \varnothing\}$.
Since $B \cap \varnothing = \varnothing$,the expression becomes $A \times \{\varnothing\}$.
This is $NOT$ equal to $\varnothing$. Therefore,the statement is False.
Corrected statement: If $A = \{1, 2\}$ and $B = \{3, 4\}$,then $A \times \{B \cap \varnothing\} = \{(1, \varnothing), (2, \varnothing)\}$.