In the diagram, not drawn to scale, A, B, C, D and E are points on the circle.
CPE and BPD are straight lines.
It is given that DC = DE, angle EPB = 104° and angle BAE = 121°.

1. Use the angle property of a cyclic quadrilateral involving points $A$, $B$, $D$ and $E$.

Since $A$, $B$, $D$ and $E$ lie on the circle, $ABDE$ is a cyclic quadrilateral.

Opposite angles in a cyclic quadrilateral are supplementary, so
$\angle BDE + \angle BAE = 180^\circ.$

$\angle BDE = 180^\circ - 121^\circ = 59^\circ.$

1. Use straight-line angles at $P$ to find the angle inside triangle $DPE$.

Since $BPD$ is a straight line,
$\angle DPE = 180^\circ - \angle EPB = 180^\circ - 104^\circ = 76^\circ.$

2. Apply the angle sum of triangle $DPE$ and use collinearity to identify $\angle DEC$.

From part (a), $\angle BDE = 59^\circ$. Since $B, P, D$ are collinear, $\angle PDE = 59^\circ$.

In $\triangle DPE$,
$\angle DEP = 180^\circ - 76^\circ - 59^\circ = 45^\circ.$
Since $C, P, E$ are collinear, $\angle DEC = \angle DEP = 45^\circ$.

1. Find the angle at $D$ in triangle $DCE$.

From part (b),
$\angle DEC=45^\circ.$
Since $DC=DE$, triangle $DCE$ is isosceles, so
$\angle DCE=\angle DEC=45^\circ.$
Hence
$\angle CDE=180^\circ-45^\circ-45^\circ=90^\circ.$

2. Use the circle theorem to prove that $CE$ is a diameter.

Since $\angle CDE=90^\circ$ and $C,D,E$ are points on the circle, the chord $CE$ subtends a right angle at the circumference.

Therefore, by the converse of the angle in a semicircle theorem, $CE$ is the diameter of the circle.

1. Use the diameter result to find the right angle at the circumference, then subtract $\angle BDE$.

Since $CE$ is a diameter, $\angle CDE=90^\circ$.

From part (a), $\angle BDE=59^\circ$.

Hence
$\angle CDB=\angle CDE-\angle BDE=90^\circ-59^\circ=31^\circ.$

1. Find the angle at $E$ in triangle $BCE$ using the diameter result and angles in the same segment.

Since $CE$ is a diameter, $\angle CBE=90^\circ$.

From part (d), $\angle CDB=31^\circ$.

Angles in the same segment give
$\angle CEB=\angle CDB=31^\circ$
would be incorrect because they do not subtend the same chord. Instead, in $\triangle CBE$,
$\angle BCE=\angle BDE=59^\circ$
(angles in the same segment, subtending chord $BE$).

Thus
$\angle BEC=180^\circ-90^\circ-59^\circ=31^\circ.$

2. Use collinearity of $C,P,E$ to identify the required angle.

Since $C,P,E$ are collinear, $EP$ lies along $EC$.

Therefore
$\angle BEP=\angle BEC=31^\circ.$

1. Find the actual value of $\angle BED$ using earlier results and collinearity.

Since $C$, $P$ and $E$ are collinear,
$\angle PED=\angle DEC=45^\circ$.

From part (e), $\angle BEP=59^\circ$.

Hence
$\angle BED=\angle BEP+\angle PED=59^\circ+45^\circ=104^\circ.$

2. Explain the contradiction if $P$ were the centre.

If $P$ were the centre, then $BD$ would pass through the centre because $B$, $P$ and $D$ are collinear. Therefore $BD$ would be a diameter.

An angle in a semicircle is $90^\circ$, so $\angle BED$ would have to be $90^\circ$.

But $\angle BED=104^\circ$, which is a contradiction. Therefore $P$ cannot be the centre of the circle.

1. Identify the angles in triangle $DPE$ needed for the sine rule.

From earlier parts, $\angle BDE=59^\circ$ and $\angle DEC=45^\circ$.

Since $B,P,D$ are collinear and $C,P,E$ are collinear,
$\angle PDE=59^\circ, \qquad \angle DEP=45^\circ.$

2. Apply the sine rule in triangle $DPE$ to find $DP$.

In $\triangle DPE$,
$\frac{DP}{\sin 45^\circ}=\frac{EP}{\sin 59^\circ}.$

Given $EP=9$,
$DP=\frac{9\sin45^\circ}{\sin59^\circ}=7.42\text{ cm} \;\text{(3 s.f.)}.$