A Point dipole $$\vec{p} = -p_0\hat{x}$$ is kept at the origin. The potential and electric field due to this dipole on the y-axis at a distance d are, respectively: (Taken V = 0 at infinity)

We are given a point electric dipole placed at the origin. Its dipole-moment vector is stated as $$\vec p=-p_0\hat x,$$ that is, its magnitude is $$|\vec p|=p_0$$ and it points along the negative x-direction.

We want the electrostatic potential $$V$$ and the electric field $$\vec E$$ at a point that lies on the y-axis at a distance $$d$$ from the origin. In Cartesian coordinates such a point has the position vector

$$\vec r=(0,d,0).$$

First let us find the potential. The standard formula for the potential due to a point dipole, with the reference $$V=0$$ at infinity, is

$$V(\vec r)=\frac{1}{4\pi\varepsilon_0}\,\frac{\vec p\cdot\vec r}{r^{3}},$$

where $$r=|\vec r|$$ is the magnitude of the position vector. Substituting the vectors involved:

$$\vec p\cdot\vec r=(-p_0\hat x)\cdot(0\,\hat x+d\,\hat y+0\,\hat z) =(-p_0)(0)+0(d)+0=0.$$

Therefore

$$V=\frac{1}{4\pi\varepsilon_0}\,\frac{0}{r^{3}}=0.$$

So the potential on the y-axis is identically zero.

Next we determine the electric field. The standard vector expression for the field of a point dipole is

$$\vec E(\vec r)=\frac{1}{4\pi\varepsilon_0}\left[\frac{3(\vec p\cdot\hat r)\hat r-\vec p}{r^{3}}\right],$$

where $$\hat r=\dfrac{\vec r}{r}$$ is the unit vector in the radial direction.

We already have $$\vec r=(0,d,0),\quad r=d,\quad\hat r=\frac{\vec r}{r}=(0,1,0).$$ Compute the scalar product that appears in the formula:

$$\vec p\cdot\hat r=(-p_0\hat x)\cdot(0\hat x+1\hat y+0\hat z)=(-p_0)(0)=0.$$

Since this dot product is zero, the first term inside the square brackets vanishes. Thus only the second term remains:

$$\vec E=\frac{1}{4\pi\varepsilon_0}\left[\frac{-\vec p}{r^{3}}\right] =-\frac{\vec p}{4\pi\varepsilon_0\,r^{3}}.$$

Replacing $$\vec p=-p_0\hat x$$ and $$r=d$$ we obtain

$$\vec E=-\frac{(-p_0\hat x)}{4\pi\varepsilon_0\,d^{3}} =\frac{p_0\hat x}{4\pi\varepsilon_0\,d^{3}}.$$

This result can also be written compactly as

$$\vec E=\frac{-\vec p}{4\pi\varepsilon_0\,d^{3}}.$$

Combining our results:

$$V=0,\qquad\vec E=\frac{-\vec p}{4\pi\varepsilon_0\,d^{3}}.$$

Comparing these expressions with the given choices, we see that they exactly match Option D.

Hence, the correct answer is Option D.