In a circle with centre O, an arc ABC subtends an angle of 138° at the centre of the circle. The chord AB is produced to a point P. Then, the measure of ∠CBP is:

To find the measure of $\angle CBP$, we can use the properties of angles subtended by an arc of a circle and cyclic quadrilaterals. Let us break down the solution step-by-step:

1. Angle subtended at the circumference:
The arc $ABC$ subtends an angle of $138°$ at the centre $O$. That is, $\angle AOC=138°$.
According to the circle theorem, the angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle.
Let $D$ be a point on the major arc $AC$ of the circle. The angle subtended by arc $ABC$ at point $D$ is $\angle ADC$.
Therefore, we have:
$\angle ADC=\frac{1}{2}\times \angle AOC$
$\angle ADC=\frac{1}{2}\times 138°=69°$

2. Cyclic Quadrilateral Property:
Since the points $A$, $B$, $C$, and $D$ lie on the circle, the quadrilateral $ABCD$ is a cyclic quadrilateral.
A key property of a cyclic quadrilateral is that the exterior angle formed by producing one of its sides is equal to the interior opposite angle.
Here, the side $AB$ is produced to the point $P$, making $\angle CBP$ the exterior angle at vertex $B$.
The interior opposite angle to $\angle CBP$ is $\angle ADC$.

Thus, we have:
$\angle CBP=\angle ADC=69°$

Therefore, the measure of $\angle CBP$ is $69°$.