3. As shown in Figure 2, the inverse proportion function $y=\frac{k}{x}(x>0)$
passes through the midpoint of the diagonal $OB$ of rectangle $OABC$ with an area of 4.
$P$ is any point on the function's graph. Then the minimum value of $OP$ is ( ).
(A) 1
(B) $\sqrt{2}$
(C) $\sqrt{3}$
(D) 2
3. B.

It is known that $k=x y=1$.
Let point $P(x, y)$. Then
$$
O P^{2}=x^{2}+y^{2} \geqslant 2 x y=2 \Rightarrow O P \geqslant \sqrt{2}
$$

Thus, the minimum value of $O P$ is $\sqrt{2}$.