We will first look at probability of $O$ not falling in any of the $3$ circles.
As we are taking all $3$ segments as diameter of $3$ circles, their radii (say, $r$ for any one of them) has to be less than the perp distance from $O$ to the segment (say, $h$) for $O$ not to be in any of the circles.
Now, $r^2 + h^2 = 1^2$ given unit circle. As the radius has to be less than the distance from $O$ to the segment, $r \leq h \implies r \leq \frac{1}{\sqrt 2} (\leq 45^0$ angle).
So the segment should subtend an angle of $\leq \frac{\pi}{2}$ at $O$.
If we take point $P$ randomly on the unit circle, the probability of the next point $Q$ being within angle $\frac{\pi}{2}$ to it (on either side of it) will be $\frac{\pi}{2\pi} = \frac{1}{2}$.
Say points $P$ and $Q$ subtend an angle of $\theta (\leq \frac{\pi}{2})$ at $O$. Then point $R$ can be within $(\frac{\pi}{2} - \theta)$ angle on either side of $P$ or $Q$. But it can also fall between point $P$ and $Q$ hence the probability of point $R$ satisfying the condition such that point $O$ is not within the $3$ circles $= \displaystyle \frac{2(\frac{\pi}{2} - \theta) + \theta}{2\pi} = \frac{\pi - \theta}{2\pi}$ where $0 \lt \theta \leq \frac{\pi}{2}$.
The probability for point $R = \frac{1}{\pi/2} \int_0^{\pi/2} (\frac{1}{2} - \frac{\theta}{2\pi}) d\theta = \frac{2}{\pi} (\frac{\pi}{4} - \frac{\pi}{16}) = \frac{3}{8}$.
So probability that none of the circles have point $O = \frac{1}{2} \times \frac{3}{8} = \frac{3}{16}$.
So desired probability $ = 1 - \frac{3}{16} = \frac{13}{16}$
