1. Given condition

We have two non-zero real matrices $A$ and $B$ of order $3 \times 3$ such that $AB=O,$ where $O$ is the zero matrix.

We must determine which option is necessarily true.

2. Key idea: if $A$ were invertible

Suppose $A$ is invertible. Then multiplying $AB=O$ on the left by $A^{-1}$ gives $A^{-1}AB=A^{-1}O \implies B=O.$ But this contradicts the fact that $B$ is a non-zero matrix.

Hence, $A$ is not invertible.

So, $\det(A)=0.$

3. Interpretation for the homogeneous system $AX=0$

Since $A$ is singular, the linear system $AX=0$ has a non-trivial solution.

Because it is a homogeneous system in $3$ variables, it always has the trivial solution $X=0$. And since $A$ is singular, there are non-trivial solutions as well.

Therefore, the system has infinitely many solutions.

So Option B is true.

4. Check the remaining options

Option A: "$AX=0$ has a unique solution"

A homogeneous system has a unique solution only when $A$ is invertible. But we proved $A$ is not invertible.

So A is false.

Option C: "$B$ is an invertible matrix"

If $B$ were invertible, then from $AB=O$ multiplying on the right by $B^{-1}$ would give $A=O,$ contradicting that $A$ is non-zero.

So C is false.

Option D: "$\operatorname{adj}(A)$ is an invertible matrix"

Since $A$ is not invertible and $A$ is $3\times 3$ with $\det(A)=0$, we use $A\,\operatorname{adj}(A)=\det(A)I=O.$ If $\operatorname{adj}(A)$ were invertible, then multiplying by its inverse would imply $A=O,$ which is impossible since $A$ is non-zero.

So D is false.

5. Conclusion

The only correct option is $\boxed{\text{B}}.$

This matches the stored correct answer.