All sides of the convex pentagon $ ABCDE$ are of equal length, and $ \angle A \equal{} \angle B \equal{} 90^{\circ}$.  What is the degree measure of $ \angle E$?

$ \textbf{(A)}\ 90 \qquad \textbf{(B)}\ 108 \qquad \textbf{(C)}\ 120 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 150$
1. We start by noting that all sides of the convex pentagon \(ABCDE\) are of equal length, and \(\angle A = \angle B = 90^\circ\). This implies that \(AB\) and \(BC\) are perpendicular to each other, forming a right angle at \(B\).

2. Since \(ABCDE\) is a convex pentagon with equal side lengths, we can infer that \(AB = BC = CD = DE = EA\).

3. Given that \(\angle A = \angle B = 90^\circ\), we can visualize the pentagon as having a square \(ABCE\) with an additional triangle \(CDE\) attached to it.

4. The quadrilateral \(ABCE\) is a square because it has four equal sides and all angles are \(90^\circ\). Therefore, \(\angle AEC = 90^\circ\).

5. The remaining part of the pentagon is the triangle \(CDE\). Since \(CD = DE = EA\), triangle \(CDE\) is equilateral, meaning all its internal angles are \(60^\circ\).

6. To find \(\angle E\), we need to consider the angles at point \(E\). Specifically, \(\angle E\) is composed of \(\angle AEC\) and \(\angle DEC\).

7. We already know that \(\angle AEC = 90^\circ\) and \(\angle DEC = 60^\circ\).

8. Therefore, \(\angle E = \angle AEC + \angle DEC = 90^\circ + 60^\circ = 150^\circ\).

The final answer is \(\boxed{150}\).