We are given: $A \cap B = A \cap C$ and $A \cup B = A \cup C$

We must determine what necessarily follows.

Use the standard idea: compare membership of an arbitrary element $x$ in $B$ and $C$.

Let $x \in B$. We will show that $x \in C$.

There are two cases:

Case 1: $x \in A$

Since $x \in A$ and $x \in B$, we get $x \in A \cap B$ But $A \cap B = A \cap C$ so $x \in A \cap C$ Hence $x \in C$.

Case 2: $x \notin A$

Since $x \in B$, we have $x \in A \cup B$ But $A \cup B = A \cup C$ so $x \in A \cup C$ Since $x \notin A$, it must be that $x \in C$.

Therefore, every element of $B$ belongs to $C$, so $B \subseteq C$

Now similarly, let $x \in C$. Again consider two cases.

Case 1: $x \in A$

Then $x \in A \cap C = A \cap B$, hence $x \in B$.

Case 2: $x \notin A$

Since $x \in C$, we have $x \in A \cup C = A \cup B$. As $x \notin A$, it follows that $x \in B$.

Therefore, $C \subseteq B$

Since both inclusions hold, $B = C$

Now check the options:

• A: $A=C$ — not necessary.
• B: $B=C$ — always true.
• C: $A \cap B = \phi$ — not necessary.
• D: $A=B$ — not necessary.

Hence the correct option is $\boxed{B}$