$K, L, M$ and $N$ are points on sides $AB, BC, CD$ and $DA$, respectively, of the unit square $ABCD$ such that $KM$ is parallel to $BC$ and $LN$ is parallel to $AB$. The perimeter of triangle $KLB$ is equal to $1$. What is the area of triangle $MND$?
1. Given a unit square \(ABCD\) with points \(K, L, M,\) and \(N\) on sides \(AB, BC, CD,\) and \(DA\) respectively, such that \(KM \parallel BC\) and \(LN \parallel AB\). The perimeter of triangle \(KLB\) is 1.
2. Let \(A = (0,0)\), \(B = (1,0)\), \(C = (1,1)\), and \(D = (0,1)\). Assume \(K = (a,0)\), \(L = (1,b)\), \(M = (c,1)\), and \(N = (0,d)\).
3. Since \(KM \parallel BC\), \(M\) must have the same \(x\)-coordinate as \(K\), so \(M = (a,1)\). Similarly, since \(LN \parallel AB\), \(N\) must have the same \(y\)-coordinate as \(L\), so \(N = (0,b)\).
4. The perimeter of \(\triangle KLB\) is given by:
   \[
   \text{Perimeter} = KB + BL + KL = 1
   \]
5. Calculate the lengths:
   \[
   KB = \sqrt{(1-a)^2 + 0^2} = 1 - a
   \]
   \[
   BL = \sqrt{(1-1)^2 + (b-0)^2} = b
   \]
   \[
   KL = \sqrt{(1-a-1)^2 + (b-0)^2} = \sqrt{(1-a)^2 + b^2}
   \]
6. Therefore, the perimeter equation is:
   \[
   (1 - a) + b + \sqrt{(1-a)^2 + b^2} = 1
   \]
7. Simplify the equation:
   \[
   \sqrt{(1-a)^2 + b^2} = a - b
   \]
8. Square both sides:
   \[
   (1-a)^2 + b^2 = (a - b)^2
   \]
9. Expand and simplify:
   \[
   1 - 2a + a^2 + b^2 = a^2 - 2ab + b^2
   \]
   \[
   1 - 2a = -2ab
   \]
   \[
   1 = 2a(1 - b)
   \]
   \[
   a = \frac{1}{2(1 - b)}
   \]
10. Since \(a\) and \(b\) must be between 0 and 1, solve for \(a\) and \(b\):
    \[
    a = \frac{1}{2(1 - b)}
    \]
11. Now, calculate the area of \(\triangle MND\):
    \[
    M = (a,1), N = (0,b), D = (0,1)
    \]
12. The area of \(\triangle MND\) is given by:
    \[
    \text{Area} = \frac{1}{2} \left| a(1 - b) \right|
    \]
13. Substitute \(a = \frac{1}{2(1 - b)}\):
    \[
    \text{Area} = \frac{1}{2} \left| \frac{1}{2(1 - b)} (1 - b) \right| = \frac{1}{2} \left| \frac{1}{2} \right| = \frac{1}{4}
    \]

The final answer is \(\boxed{\frac{1}{4}}\).