$x(t)=A \sin (\omega t-k x)+A \sin (\omega t+k x+\delta)$

$x(t)=2 A \cos (k x) \sin \left(\omega t+\frac{\delta}{2}\right)$

For all particles to be simultaneously at rest, the value of the sine function must be equal to $zero$.

$\text { i.e. } \omega t +\frac{\delta}{2}= n \pi$

$\Rightarrow t =\frac{1}{\omega}\left( n \pi-\frac{\delta}{2}\right)$

$\omega=\frac{2 \pi}{ T }$

$\Rightarrow t =\frac{ T }{2 \pi}\left( n \pi-\frac{\delta}{2}\right)$

$t _1=\frac{ T }{2 \pi}\left( n \pi-\frac{\delta}{2}\right)$

$t _2=\frac{ T }{2 \pi}\left(( n +1) \pi-\frac{\delta}{2}\right)$

$t _2- t _1=\frac{ T }{2}$

So, the time between this event happening twice is half to time period, so in one cycle this would happen twice. So Option $(A)$

For the particle to be at positive extreme the sine function can take a value of 1 only.

It can be shown that this happens at an interval of ' $T$ '

So it will happen twice in a time period if the displacement is max at the start of the time period and once more at the end of the time period,

or else it would happen only once in a time period. Hence Option $(C)$