Since the wave is an transverse wave the particles will move perpendicular to the direction of propagation so we can eliminate B,D

Now we know that the velocity of any particle is given by V(Particle)=-V(wave)*(Slope of the graph at particle)

Proof: Let equation of wave be y=ASin(wt-kx+a) Then Velocity of particle is: dy/dt= V(p)=AwCos(wt-kx) -i

Slope of graph is:dy/dx=m=-AkCos(wt-kx+a) -ii

From equation i,ii : V(p)=-w/k(m)=-V(w)*m {Since Velocity of wave is given by w/k}

As per the given question wave travels from left to right I.e v(w)=+ve and the Slope of graph at given particle too is +ve So V(p)=(-ve)(+ve)(+ve)=-ve so it moves downward

Since the wave is an transverse wave the particles will move perpendicular to the direction of propagation so we can eliminate B,D

Now we know that the velocity of any particle is given by $v_{Particle}=-v_{wave}*\text{(Slope of the graph at particle)}$

Proof: Let equation of wave be $y=A\;sin(wt-kx+a)$ then velocity of particle is: $\frac{dy}{dt} = v_{particle}=A \omega \; cos(wt-kx)$ ①

Slope of graph is: $dy/dx= m =- Ak \; cos(wt-kx+a)$ ②

From equation ①,②: $v_{particle}=-\omega/k*m=-v_{wave}*m$ (Since velocity of wave is given by $\frac{w}{k}$)

As per the given question wave travels from left to right ie v(w)= positive and the slope of graph at given particle location is also positive So if we apply this, $-(\text{positive }v_{wave})(\text{positive slope})=\text{negative }v_{particle}$, so it moves downward (c is your answer)