Explanation

(d) As is clear from figure, at$t = 0,x = 0$, displacement$y = 0$. Therefore, option (a) or (d)may be correct.
In case of (d);$y = A\sin{(kx - \omega t)}$
$\frac{dy}{dt} = A\cos{(kx - \omega t)\lbrack - \omega\rbrack}$
$\frac{dy}{dx} = A\cos{(kx - \omega t)\lbrack k\rbrack}$
$\frac{\frac{dy}{dt}}{dy/dx} = \frac{- \omega A\cos{(kx - \omega t)}}{kA\cos{(kx - \omega t)}} = - \frac{\omega}{k} = - v$
$\frac{dy}{dt} = - v\left( \frac{dy}{dx} \right)$
$\ ie\ particle\ velocity = - (wave\ speed) \times slope$.
And slope at $x = 0$ and $t = 0$is positive, in figure. Therefore, particle velocity is in negative y-direction.