We are given a cyclic quadrilateral $ABCD$ inscribed in a circle. We know that $\angle DBC = 47^\circ$ and $\angle ADB = 28^\circ$. We want to find the measure of the angle $\angle DCP$.

1. Problem Description

We are given a cyclic quadrilateral $ABCD$ inscribed in a circle. We know that $\angle DBC = 47^\circ$ and $\angle ADB = 28^\circ$. We want to find the measure of the angle $\angle DCP$.

2. Solution Steps

Since $ABCD$ is a cyclic quadrilateral, the vertices $A$, $B$, $C$, and $D$ lie on the circumference of the circle.

Since angles subtended by the same chord are equal, we have $\angle DAC = \angle DBC = 47^\circ$.

Also, $\angle ACB = \angle ADB = 28^\circ$.

We know that the sum of angles in $\triangle ADC$ is $180^\circ$. Therefore, $\angle ACD = 180^\circ - \angle DAC - \angle ADC = 180^\circ - 47^\circ - 28^\circ = 105^\circ$.

Since $ABCD$ is a cyclic quadrilateral, we have $\angle ABC + \angle ADC = 180^\circ$, and $\angle BAD + \angle BCD = 180^\circ$.

We have $\angle BCD = \angle BCA + \angle ACD = 28^\circ + 105^\circ = 133^\circ$.

Since $CDP$ is a straight line, $\angle DCP + \angle BCD = 180^\circ$.

$\angle DCP = 180^\circ - \angle BCD$.

We can also write $\angle DCP = \angle DAB$.

$\angle DAB = \angle DAC + \angle CAB$.

Also, $\angle CAB = \angle CDB$.

We are given $\angle DBC = 47^\circ$ and $\angle ADB = 28^\circ$.

Since $\angle CAB = \angle CDB$, and quadrilateral $ABCD$ is cyclic, we have $\angle CDB = \angle CAB$. Also $\angle CAB = \angle CDB = 28^\circ$.

So $\angle DAB = \angle DAC + \angle CAB = 47^\circ + 28^\circ = 75^\circ$.

Since $\angle DCP = 180^\circ - \angle BCD$, and $\angle BCD + \angle DAB = 180^\circ$, then $\angle DCP = \angle DAB = 75^\circ$.

3. Final Answer

75 degrees