For each real number $a$ with $0 \leq a \leq 1$, let numbers $x$ and $y$ be chosen independently at random from the intervals $[0, a]$ and $[0, 1]$, respectively, and let $P(a)$ be the probability that
\[\sin^2{(\pi x)} + \sin^2{(\pi y)} > 1\]
What is the maximum value of $P(a)?$
$\textbf{(A)}\ \frac{7}{12} \qquad\textbf{(B)}\ 2 - \sqrt{2} \qquad\textbf{(C)}\ \frac{1+\sqrt{2}}{4} \qquad\textbf{(D)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(E)}\ \frac{5}{8}$
Let's start first by manipulating the given inequality.
\[\sin^{2}{(\pi x)}+\sin^{2}{(\pi y)}>1\]
\[\sin^{2}{(\pi x)}>1-\sin^{2}{(\pi y)}=\cos^{2}{(\pi y)}\]
Let's consider the boundary cases: $\sin{(\pi x)}=\pm \cos{(\pi y)}$.
\[\sin{(\pi x)}=\pm \cos{(\pi y)}=\sin{(\tfrac 12 {\pi}\pm \pi y)}\]
Solving the first case gives us \[y=\tfrac{1}{2}-x \quad \textrm{and} \quad  y=x-\tfrac{1}{2}.\] Solving the second case gives us \[y=x+\tfrac{1}{2}\quad \textrm{and} \quad y=\tfrac{3}{2}-x.\] If we graph these equations in $[0,1]\times[0,1]$, we get a rhombus shape. 

Testing points in each section tells us that the inside of the rhombus satisfies the inequality in the problem statement.
From the region graph, notice that in order to maximize $P(a)$, $a\geq\tfrac{1}{2}$. We can solve the rest with geometric probability.
Instead of maximizing $P(a)$, we minimize $Q(a)=1-P(a)$. $Q(a)$ consists of two squares (each broken into two triangles), one of area $\tfrac{1}{4}$ and another of area $(a-\tfrac 12)^2$. To calculate $Q(a)$, we divide this area by $a$, so \[Q(a) = \frac{1}{a}\left(\frac{1}{4}+(a-\tfrac 12)^2\right) = \frac{1}{a}\left(\frac{1}{2}+a^2-a\right)= a+\frac 1{2a}-1.\]
By AM-GM, $a+\frac{1}{2a}\geq 2\sqrt{\frac{a}{2a}}=\sqrt{2}$, which we can achieve by setting $a=\frac{\sqrt{2}}{2}$.
Therefore, the maximum value of $P(a)$ is $1-\min(Q(a))$, which is $1-(\sqrt{2}-1) =\boxed{\textbf{(B)}\ 2 - \sqrt{2}}$