Let $ABCD$ be a convex quadrilateral such that $m \left (\widehat{ADB} \right)=15^{\circ}$, $m \left (\widehat{BCD} \right)=90^{\circ}$. The diagonals of quadrilateral are perpendicular at $E$. Let $P$ be a point on $|AE|$ such that $|EC|=4, |EA|=8$ and $|EP|=2$. What is $m \left (\widehat{PBD} \right)$?

$ 
\textbf{(A)}\ 15^{\circ}
\qquad\textbf{(B)}\ 30^{\circ}
\qquad\textbf{(C)}\ 45^{\circ}
\qquad\textbf{(D)}\ 60^{\circ}
\qquad\textbf{(E)}\ 75^{\circ}
$
1. Given that $ABCD$ is a convex quadrilateral with $m(\widehat{ADB}) = 15^\circ$ and $m(\widehat{BCD}) = 90^\circ$, and the diagonals intersect perpendicularly at $E$.
2. We are given the lengths $|EC| = 4$, $|EA| = 8$, and $|EP| = 2$.
3. We need to find $m(\widehat{PBD})$.

First, let's use the given information about the lengths and the angles to find the necessary relationships.

4. Since $m(\widehat{BCD}) = 90^\circ$, triangle $BCD$ is a right triangle with $E$ as the foot of the altitude from $B$ to $CD$.
5. The diagonals intersect perpendicularly at $E$, so $BE \cdot DE = CE^2$ (this is a property of the right triangle where the altitude to the hypotenuse creates two similar triangles).

6. Given $|EC| = 4$, we have:
   \[
   BE \cdot DE = 4^2 = 16
   \]

7. Since $|EA| = 8$ and $|EP| = 2$, we can use the Power of a Point theorem at point $P$:
   \[
   PE \cdot EA = 2 \cdot 8 = 16
   \]

8. This implies that $\triangle BEP \sim \triangle AED$ by the AA (Angle-Angle) similarity criterion because they share the angle at $E$ and the right angle at $E$.

9. From the similarity, we have:
   \[
   \angle PBE = \angle DAE
   \]

10. Given $m(\widehat{ADB}) = 15^\circ$, we know that:
    \[
    \angle DAE = 90^\circ - 15^\circ = 75^\circ
    \]

11. Therefore, $\angle PBE = 75^\circ$.

12. Finally, since $\angle PBD = \angle PBE - \angle EBD$ and $\angle EBD = 0^\circ$ (as $E$ lies on $BD$), we have:
    \[
    \angle PBD = 75^\circ
    \]

Conclusion:
\[
\boxed{75^\circ}
\]