The question is a little confusing but let me try to put some information in a way that hopefully will answer the question.
I assume that the line is a solid wore or a thin rod.
In such a case every particle on the rod will posses linear velocity $v$ that reduces as you move from point P towards the center. The center actually has zero linear velocity (visualize the wing of a fan, you can trace the movement of inner edge of the wing but not the outer. The reason is that the velocity of outer edge is more than that of the inner hence difficult to track by eyes). The other way of interpreting is that the outer edge has to cover a larger circumference (owing to larger radius) than the inner edge (smaller radius) in same time $t$. Thus center that has zero radius has zero velocity.
However, the angular velocity $\omega$ measures the angular sweep of the wire OP per unit time. You can clearly see that every point on OP sweeps a similar angle and hence the angular velocity of every point on OP is the same.
So the angular velocity of the wire and its every point is same but the linear velocity reduces as you move from P towards O.
The relation between linear velocity v and angular velocity is given by the equation
$v = \omega r$
Mathematically also since $\omega$ is constant, $v$ must increase as $r$ increases and reduce as $r$ reduces
Use this note to figure out any answer around rotational motion of OP
You may like to watch this video on the topic made by me for better understanding
Rotation - Angular displacement, Velocity and acceleration