Answer
The greatest value of $x$ at which the light waves arrive exactly out of phase is $1.7\mu\mathrm{m}$.
Work Step by Step
At $x={x_s\over 2}=5\times 10^{-7}$, according to the graph, the phase difference between the two waves is $\phi=4\pi$.
This translates into a path difference of $\Delta L={\lambda\over 2\pi}\phi={\lambda\over 2\pi}4\pi=2\lambda$
Using Pythagoras theorem,
$\Delta \lambda=\sqrt{x^2+d^2}-x$
For $x=x_s/2=x’$
\begin{align*}
\Delta \lambda&=\sqrt{x’^2+d^2}-x’\\
2\lambda&=\sqrt{x’^2+d^2}-x’\\
(2\lambda+x’)^2&=x’^2+d^2\\
4\lambda^2+x’^2+4\lambda x’&=x'^2+d^2\\
d^2&=4\lambda^2+4\lambda x’\\
d^2&=4\lambda(\lambda+x’)
\end{align*}
If light waves arriving at P must be exactly out of phase, they must interfere destructively. I.e.,
$\Delta L=\left(m+{1\over 2}\right)\lambda$
Again using the expression for the path difference for the required distance $x$ of P from $S_1$,
\begin{align*}
\sqrt{x^2+d^2}-x&=\left(m+{1\over 2}\right)\lambda\\
d^2+x^2&=\left(\left(m+{1\over 2}\right)\lambda+x\right)^2\\
d^2+x^2&=\left(m+{1\over 2}\right)^2\lambda^2+x^2+2x\left(m+{1\over 2}\right)\lambda\\
d^2&={(2m+1)^2\over 4}\lambda^2+2x\left(m+{1\over 2}\right)\lambda\\
2x\left(m+{1\over 2}\right)\lambda&=d^2-{(2m+1)^2\over 4}\lambda^2\\
x&={d^2\over \lambda (2m+1)}-{2m+1\over 4}\lambda
\end{align*}
Substituting the expression for $d^2$ derived earlier,
\begin{align*}
x={4\lambda(\lambda+x’)\over \lambda (2m+1)}-{2m+1\over 4}\lambda\\
x={4(\lambda+x’)\over 2m+1}-{2m+1\over 4}\lambda
\end{align*}
To obtain the greatest value of x, we set $m=0$ (first dark fringe). Then
$x=4(\lambda+x’)-{\lambda\over 4}$
Substituting the given values, $\lambda=400\times 10^{-9}\mathrm{m}$ and $x’=x_s/2=5\times 10^{-7}\mathrm{m}$, we get,
\begin{align*}
x&=4(\lambda+x’)-{\lambda\over 4}\\
x&=4(400\times 10^{-9}+5\times 10^{-10})-{400\times 10^{-9}\over 4}\\
&=1800\times 10^{-9}-100\times 10^{-9}\\
&=1700\times 10^{-9}\\
&=1.7\mu\mathrm{m}
\end{align*}
Hence the greatest value of $x$ at which the light waves arrive exactly out of phase is $1.7\mu\mathrm{m}$.
