Let the locus of the mid-point of the chord through the origin O of the parabola $$y^{2}= 4x$$ be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the ratio 3 :1, is:

the standard form $$y^{2}=4ax$$ gives $$a=1$$.

Equation of a chord of the parabola whose mid-point is $$(h,k)$$.
For $$y^{2}=4ax$$ the formula for the chord with mid-point $$(h,k)$$ is $$k\,y-2a\,(x+h)=k^{2}-4a\,h \qquad -(1)$$ (This is the result $$T=S_{1}$$ for a parabola.)

Condition that the chord passes through the origin $$O(0,0)$$.
Substitute $$x=0,\;y=0$$ in $$(1)$$: $$0\cdot k-2a(0+h)=k^{2}-4a\,h$$ $$-2a\,h=k^{2}-4a\,h$$ $$k^{2}=2a\,h \qquad -(2)$$

With $$a=1$$, (2) becomes $$k^{2}=2h$$.
Hence the locus of the mid-point $$P(h,k)$$ of every chord through the origin is $$y^{2}=2x$$.
This curve is denoted by $$S$$.

Coordinates of the point which divides $$OP$$ internally in the ratio $$3:1$$.
For points $$O(0,0)$$ and $$P(h,k)$$, the section point $$G(x,y)$$ with $$\frac{OG}{GP}=3:1$$ is $$x=\frac{3h+1\cdot 0}{3+1}=\frac{3h}{4}, \quad y=\frac{3k+1\cdot 0}{3+1}=\frac{3k}{4} \qquad -(3)$$

$$h=\frac{4x}{3},\;k=\frac{4y}{3}$$.
Insert $$h=\frac{4x}{3},\;k=\frac{4y}{3}$$ into the relation $$k^{2}=2h$$: $$\left(\frac{4y}{3}\right)^{2}=2\left(\frac{4x}{3}\right)$$ $$\frac{16y^{2}}{9}=\frac{8x}{3}$$  

$$2y^{2}=3x$$.

the locus of the point dividing $$OP$$ in the ratio $$3:1$$ is $$\boxed{2y^{2}=3x}$$, which corresponds to Option B.