Let $k,n$ be positive integers. There are $kn$ points $P_1, P_2, \cdots, P_{kn}$ on a circle. We can color each points with one of color $ c_1, c_2, \cdots , c_k $. In how many ways we can color the points satisfying the following conditions?

(a) Each color is used $ n $ times.

(b) $ \forall  i \not = j $, if $ P_a $ and $ P_b $ is colored with color $ c_i $ , and $ P_c $ and $ P_d $ is colored with color $ c_j $, then the segment $ P_a P_b $ and segment $ P_c P_d $ doesn't meet together.
1. **Understanding the Problem:**
   We need to color \( kn \) points on a circle using \( k \) colors such that each color is used exactly \( n \) times, and no two chords of the same color intersect.

2. **Initial Setup:**
   Suppose \( P_aP_b \) is a chord between two points colored \( c \). Let \( p_1, p_2, \ldots, p_r \) be the points between \( P_a \) and \( P_b \) on the circle. If for some \( i \), point \( p_i \) is colored \( c' \neq c \), then all points colored \( c' \) must lie between \( P_a \) and \( P_b \), else we would contradict the non-intersecting condition.

3. **Coloring Strategy:**
   If \( P_1, \ldots, P_j \) contain the colors \( c_1, \ldots, c_s \) and \( P_{j+1} \) is colored \( c_{s+1} \), then all \( n \) points colored \( c_{s+1} \) must appear before the next point with a color in \( \{c_1, \ldots, c_s\} \).

4. **Marking Points:**
   Consider an admissible coloring of the points \( P_1, \ldots, P_{nk} \). Moving clockwise from \( P_1 \), mark the first point at which each color appears. We mark \( k \) points \( \{P_1, P_{i_2}, \ldots, P_{i_k}\} \). This set uniquely determines the coloring (up to the order of the colors).

5. **Coloring Process:**
   - Color \( P_1 \) with \( c_1 \).
   - \( P_2, \ldots, P_{i_2-1} \) are all colored \( c_1 \) because that's the only color available.
   - \( i_2 \le n+1 \) because we can color at most \( n \) points \( c_1 \).
   - Color \( P_{i_2} \) with \( c_2 \). By the previous condition, the only color we can use now is \( c_2 \). So \( P_{i_2}, P_{i_2+1}, \ldots, P_{i_3-1} \) are all \( c_2 \).
   - Continue this process: every time we hit a marked point \( P_{i_j} \), we use a new color. After we have colored \( n \) points \( c_k \), we revert back to the color we were using previous to \( c_k \) (unless it has also been used \( n \) times, in which case we revert back further).

6. **Counting the Marked Points:**
   - We need to count the number of sets \( \{P_1, P_{i_2}, \ldots, P_k\} \) and the order in which the colors appear (which is \( k! \) ways).
   - We must have \( i_j \le n(j-1) + 1 \) else we will run out of colors to revert back to before we hit \( P_{i_j} \).
   - The first term is always \( P_1 \), so we count the number of sequences \( \{i_2, \ldots, i_k\} \) such that \( 1 < i_2 < i_3 < \ldots < i_k \) and \( i_j \le n(j-1) + 1 \).

7. **Counting Sequences:**
   - Let \( R(n, k-1) \) be the number of such sequences.
   - Break the interval \( [2, n(k-1)+1] \) into \( k-1 \) blocks of \( n \) consecutive integers and count the number of marked points \( i_j \) in each block.
   - Let \( \mathcal{B}_{k-1} \) be the family of \( (k-1) \)-tuples \( (b_1, \ldots, b_{k-1}) \) such that \( b_1 + \ldots + b_j \le j \) for each \( j = 1, 2, \ldots, k-2 \) and \( b_1 + \ldots + b_{k-1} = k-1 \).
   - Then \( R(n, k-1) = \sum_{B \in \mathcal{B}_{k-1}} \prod_{i=1}^{k-1} \binom{n}{b_i} \).

8. **Final Calculation:**
   - The final answer is \( k! R(n, k-1) \).

The final answer is \( \boxed{ k! R(n, k-1) } \).