In fact one can prove a stronger result -- namely the probability is bounded away from zero, so long as $B \ge (\log N)^\delta$ for some $\delta >0$. This is best possible, by taking $N$ to be the product of the first few primes. Since $\phi(N/d) \ge \phi(N)/d$, our probability is bounded below by
$$
\frac{\phi(N)}{N} \sum_{d|N, d\le B} \frac{1}{d}\ge \frac{\phi(N)}{N}\sum_{d|N, d\le B} \frac{\mu(d)^2}{d},
$$
where in the last sum we have restricted attention to square-free $d$ for simplicity.

Now I claim that for large enough $B$ (and any $N$, independent of $B$) one has
$$
\sum_{d|N, d\le B} \frac{\mu(d)^2}{d} \ge \frac{1}{4} \prod_{p|N, p\le \sqrt{B}} \Big(1+\frac 1p\Big). \tag{1}
$$
Assuming the claim, we get a lower bound for our probability of
$$
\ge \frac 14 \frac{\phi(N)}{N} \prod_{p|N, p\le \sqrt{B}} \Big(1+\frac 1p\Big)
\ge \frac{1}{4} \prod_{p} \Big(1-\frac 1{p^2}\Big) \prod_{p|N, p>\sqrt{B}} \Big(1-\frac 1p\Big)= \frac{3}{2\pi^2} \prod_{p|N, p>\sqrt{B}} \Big(1-\frac{1}{p}\Big).
$$
If now $B\ge (\log N)^{\delta}$ then
$$
\prod_{p|N, p>\sqrt{B}} \Big(1-\frac 1p\Big) \ge \prod_{(\log N)^{\delta/2}<p <(\log N)^2}\Big(1-\frac 1p\Big) \prod_{p>(\log N)^2, p|N} \Big(1-\frac 1p\Big),
$$
and by Mertens's theorem the first factor above is $\gg \delta$, and trivially the second factor is $1+o(1)$. This completes the proof.

It remains to settle the claim (1). With $\alpha =1/\log B$ note that
$$
\sum_{d|N, d\le B} \frac{\mu(d)^2}{d} \ge \sum_{\substack{d|N, d\le B \\ p|d \implies p\le \sqrt{B}}} \frac{\mu(d)^2}{d} \ge \prod_{p|N, p\le \sqrt{B}} \Big(1+\frac 1p\Big) - B^{-\alpha} \sum_{\substack{d|N\\ p|d\implies p\le \sqrt{B} }} \frac{\mu(d)^2 d^{\alpha}}{d}.
$$
The second term above is
$$
e^{-1} \prod_{p|N, p\le \sqrt{B}} \Big(1 + \frac{p^{\alpha}}{p}\Big) \le e^{-1} \prod_{p|N, p\le \sqrt{B}} \Big(1+\frac 1p\Big) \exp\Big(\sum_{p|N, p\le \sqrt{B}} \frac{p^{\alpha}-1}{p}\Big).
$$
Now for large enough $B$, (since $(e^{t}-1)/t \le (\sqrt{e}-1)/(1/2)$ for $0\le t\le 1/2$)
$$
\sum_{p\le \sqrt{B}} \frac{p^{1/\log B}-1}{p} \le\sum_{p\le \sqrt{B}} \frac{\log p}{p\log B} \Big(\frac{\sqrt{e}-1}{1/2}\Big) = \sqrt{e}-1 +o(1),
$$
and using this above, we obtain
$$
\sum_{d|N, d\le B} \frac{\mu(d)^2}{d} \ge \prod_{p|N, p\le \sqrt{B}} \Big(1+\frac 1p\Big) \Big(1- e^{-1} (e^{\sqrt{e}-1}+o(1)) \Big) \ge \frac 14 \prod_{p|N, p\le \sqrt{B}} \Big(1+\frac 1p\Big),
$$
proving the claim (1).