5. Since $\triangle A D P$ is isosceles and $\angle A D P=2 x^{\circ}$, then $\angle D A P=\angle D P A=\frac{1}{2}\left(180^{\circ}-2 x^{\circ}\right)=90^{\circ}-x^{\circ}$. Similarly, since $\triangle C D P$ is isosceles and $\angle C D P=2 x^{\circ}$, then $\angle D C P=\angle D P C=90^{\circ}-x^{\circ}$.
Since $\triangle B A P$ is isosceles and $\angle B A P=(x+5)^{\circ}$, then $\angle B P A=\frac{1}{2}\left(180^{\circ}-(x+5)^{\circ}\right)=\frac{1}{2}(175-x)^{\circ}$.
Since the measures of the four angles around point $P$ must add to $360^{\circ}$, then
$$
\begin{aligned}
\angle D P C+\angle D P A+\angle B P A+\angle B P C & =360^{\circ} \\
\left(90^{\circ}-x^{\circ}\right)+\left(90^{\circ}-x^{\circ}\right)+\frac{1}{2}(175-x)^{\circ}+(10 x-5)^{\circ} & =360^{\circ} \\
(90-x)+(90-x)+\frac{1}{2}(175-x)+(10 x-5) & =360 \\
(180-2 x)+(180-2 x)+(175-x)+(20 x-10) & =720 \\
15 x+525 & =720 \\
15 x & =195 \\
x & =13
\end{aligned}
$$

Therefore, $x=13$.
ANSWER: 13
5. In the diagram, point $P$ is inside quadrilateral $A B C D$. Also, $D A=D P=D C$ and $A P=A B$. If $\angle A D P=\angle C D P=2 x^{\circ}, \angle B A P=(x+5)^{\circ}$, and $\angle B P C=(10 x-5)^{\circ}$, what is the value of $x$ ?