Charges are placed on the vertices of a square as shown. Let $$\vec{E}$$ be the electric field and $$V$$ the potential at the centre. If the charges on $$A$$ and $$B$$ are interchanged with those on $$D$$ and $$C$$ respectively, then

Let the centre of the square be $$O$$ and the side of the square be $$a$$. Choose Cartesian axes through $$O$$ so that the four vertices have co-ordinates

$$A\,(+a/2,+a/2),\;B\,(-a/2,+a/2),\;C\,(-a/2,-a/2),\;D\,(+a/2,-a/2).$$

The distance of every vertex from the centre is the same,

$$r = \frac{a}{\sqrt{2}}.$$ Hence all four charges are equidistant from $$O$$.

Denote the initial charges by $$q_A,\;q_B,\;q_C,\;q_D$$ placed at $$A,B,C,D$$ respectively.

Potential at the centre (before interchange)
Using the superposition principle, the electric potential at a point is the algebraic sum of the potentials due to individual charges because potential is a scalar. Therefore

$$V_{\text{old}} \;=\; k\,\frac{q_A + q_B + q_C + q_D}{r},$$ where $$k = \tfrac{1}{4\pi\varepsilon_0}.$$

Electric field at the centre (before interchange)
For any vertex the magnitude of the field at $$O$$ is

$$E_i = k\,\frac{|q_i|}{r^2} = k\,\frac{2\,q_i}{a^{2}},$$

and its direction is along the line joining that vertex to the centre (outward for a positive charge, inward for a negative one). Writing the four position vectors from $$O$$ to the vertices,

$$\vec{r}_A = \tfrac{a}{2}\,(\hat{i}+\hat{j}),\; \vec{r}_B = \tfrac{a}{2}\,(-\hat{i}+\hat{j}),\; \vec{r}_C = \tfrac{a}{2}\,(-\hat{i}-\hat{j}),\; \vec{r}_D = \tfrac{a}{2}\,(\hat{i}-\hat{j}),$$

the electric field at $$O$$ is

$$\vec{E}_{\text{old}} \;=\; k\,\frac{2}{a^{2}}\, \Bigl(q_A\,\hat{r}_A + q_B\,\hat{r}_B + q_C\,\hat{r}_C + q_D\,\hat{r}_D\Bigr),$$ where each $$\hat{r}_i$$ is the unit vector along $$\vec{r}_i$$.

Interchange of charges
The question says: “charges on $$A$$ and $$B$$ are interchanged with those on $$D$$ and $$C$$ respectively.” In symbols this means

$$q_A \longleftrightarrow q_D,\qquad q_B \longleftrightarrow q_C.$$

After interchange, the charges occupying the vertices become

$$A:D,\; B:C,\; C:B,\; D:A.$$

Potential after interchange
Because the distance $$r$$ is unchanged and the potential is a scalar sum, the new potential is

$$V_{\text{new}} = k\,\frac{q_D + q_C + q_B + q_A}{r} = k\,\frac{q_A + q_B + q_C + q_D}{r} = V_{\text{old}}.$$

Hence the electric potential at the centre does not change.

Electric field after interchange
The magnitudes are the same as before, but the charges now multiply different direction vectors:

$$\vec{E}_{\text{new}} = k\,\frac{2}{a^{2}}\, \Bigl(q_D\,\hat{r}_A + q_C\,\hat{r}_B + q_B\,\hat{r}_C + q_A\,\hat{r}_D\Bigr).$$

In general $$\vec{E}_{\text{new}}\neq\vec{E}_{\text{old}}$$ unless the pairs $$q_A=q_D$$ and $$q_B=q_C$$ already happened to be equal. Therefore the electric field changes.

Conclusion
The interchange leaves the potential unchanged but alters the electric field.

Option D which is: $$\vec{E}$$ changes, $$V$$ remains unchanged.