For the four circles $$M, N, O$$ and $$P$$, following four equations are given:
Circle M: $$x^2 + y^2 = 1$$
Circle N: $$x^2 + y^2 - 2x = 0$$
Circle O: $$x^2 + y^2 - 2x - 2y + 1 = 0$$
Circle P: $$x^2 + y^2 - 2y = 0$$
If the centre of circle M is joined with centre of circle N, further centre of circle N is joined with centre of circle O, centre of circle O is joined with the centre of circle P and lastly, centre of circle P is joined with centre of circle M, then these lines form the sides of a:

We find the centre of each circle. Circle M: $$x^2 + y^2 = 1$$ has centre $$(0, 0)$$. Circle N: $$x^2 + y^2 - 2x = 0$$ becomes $$(x-1)^2 + y^2 = 1$$, so centre is $$(1, 0)$$. Circle O: $$x^2 + y^2 - 2x - 2y + 1 = 0$$ becomes $$(x-1)^2 + (y-1)^2 = 1$$, so centre is $$(1, 1)$$. Circle P: $$x^2 + y^2 - 2y = 0$$ becomes $$x^2 + (y-1)^2 = 1$$, so centre is $$(0, 1)$$.

The four centres are $$M(0,0)$$, $$N(1,0)$$, $$O(1,1)$$, and $$P(0,1)$$. Joining them in order $$M \to N \to O \to P \to M$$ forms a quadrilateral.

All four sides have length 1: $$MN = NO = OP = PM = 1$$. The diagonals are $$MO = \sqrt{1+1} = \sqrt{2}$$ and $$NP = \sqrt{1+1} = \sqrt{2}$$, so the diagonals are equal.

Since all sides are equal and the diagonals are equal, this is a square.

The answer is Option B: Square.