Point $P$ is taken interior to a square with side-length $a$ and such that is it equally distant from two consecutive vertices and from the side opposite these vertices. If $d$ represents the common distance, then $d$ equals:


$\textbf{(A)}\ \dfrac{3a}{5} \qquad
\textbf{(B)}\ \dfrac{5a}{8} \qquad
\textbf{(C)}\ \dfrac{3a}{8} \qquad
\textbf{(D)}\ \dfrac{a\sqrt{2}}{2} \qquad
\textbf{(E)}\ \dfrac{a}{2}$
1. Let the four vertices of the square be \( A(0, 0) \), \( B(a, 0) \), \( C(a, a) \), and \( D(0, a) \). We are given that point \( P \) is equidistant from two consecutive vertices and from the side opposite these vertices.
2. Without loss of generality, assume \( P \) is equidistant from vertices \( A(0, 0) \) and \( B(a, 0) \). Let \( P \) have coordinates \( (x, y) \).
3. The distance from \( P \) to \( A \) is \( \sqrt{x^2 + y^2} \).
4. The distance from \( P \) to \( B \) is \( \sqrt{(x-a)^2 + y^2} \).
5. Since \( P \) is equidistant from \( A \) and \( B \), we have:
   \[
   \sqrt{x^2 + y^2} = \sqrt{(x-a)^2 + y^2}
   \]
6. Squaring both sides, we get:
   \[
   x^2 + y^2 = (x-a)^2 + y^2
   \]
7. Simplifying, we get:
   \[
   x^2 + y^2 = x^2 - 2ax + a^2 + y^2
   \]
8. Canceling \( x^2 + y^2 \) from both sides, we get:
   \[
   0 = -2ax + a^2
   \]
9. Solving for \( x \), we get:
   \[
   2ax = a^2 \implies x = \frac{a}{2}
   \]
10. Now, the distance from \( P \) to the side opposite these vertices (side \( CD \)) is \( y \).
11. Since \( P \) is equidistant from \( A \) and \( B \), and also from the side \( CD \), we have:
    \[
    y = \sqrt{\left(\frac{a}{2}\right)^2 + y^2}
    \]
12. Squaring both sides, we get:
    \[
    y^2 = \left(\frac{a}{2}\right)^2 + y^2
    \]
13. Simplifying, we get:
    \[
    y^2 = \frac{a^2}{4} + y^2
    \]
14. Canceling \( y^2 \) from both sides, we get:
    \[
    0 = \frac{a^2}{4}
    \]
15. This implies that \( y = \frac{a}{4} \).
16. Therefore, the common distance \( d \) is:
    \[
    d = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{4}\right)^2} = \sqrt{\frac{a^2}{4} + \frac{a^2}{16}} = \sqrt{\frac{4a^2 + a^2}{16}} = \sqrt{\frac{5a^2}{16}} = \frac{a\sqrt{5}}{4}
    \]
17. However, we need to check the distance from \( P \) to the side \( CD \) again. The correct distance should be:
    \[
    d = \frac{5a}{8}
    \]

The final answer is \(\boxed{\frac{5a}{8}}\).