Two perpendicular chords are drawn through a given interior point $P$ of a circle with radius $R.$ Determine, with proof, the maximum and the minimum of the sum of the lengths of these two chords if the distance from $P$ to the center of the circle is $kR.$
1. **Define the problem and given parameters:**
   - We have a circle with radius \( R \).
   - A point \( P \) inside the circle is at a distance \( kR \) from the center \( O \) of the circle, where \( 0 \leq k \leq 1 \).
   - Two perpendicular chords \( AB \) and \( CD \) pass through \( P \).

2. **Power of a Point Theorem:**
   - The power of point \( P \) with respect to the circle is given by:
     \[
     p_w(P) = OP^2 - R^2 = (kR)^2 - R^2 = R^2(k^2 - 1)
     \]

3. **Lengths of the Chords:**
   - Let \( AB \) and \( CD \) be the lengths of the chords passing through \( P \).
   - By the Power of a Point Theorem, the product of the segments of each chord through \( P \) is equal to the power of \( P \):
     \[
     PA \cdot PB = PC \cdot PD = R^2 - OP^2 = R^2 - (kR)^2 = R^2(1 - k^2)
     \]

4. **Sum of the Squares of the Chords:**
   - Since \( AB \) and \( CD \) are perpendicular and pass through \( P \), we can use the Pythagorean theorem in the context of the circle:
     \[
     AB^2 + CD^2 = 4(R^2 - OP^2) = 4(R^2 - k^2R^2) = 4R^2(1 - k^2)
     \]

5. **Maximum and Minimum Sum of the Chords:**
   - To find the maximum and minimum of \( AB + CD \), we use the Cauchy-Schwarz inequality:
     \[
     (AB + CD)^2 \leq 2(AB^2 + CD^2)
     \]
     Substituting the value from step 4:
     \[
     (AB + CD)^2 \leq 2 \cdot 4R^2(1 - k^2) = 8R^2(1 - k^2)
     \]
     Taking the square root of both sides:
     \[
     AB + CD \leq 2R \sqrt{2(1 - k^2)}
     \]

6. **Minimum Sum of the Chords:**
   - The minimum value of \( AB + CD \) occurs when the chords are degenerate (i.e., when \( P \) is at the center of the circle, \( k = 0 \)):
     \[
     AB + CD = 0
     \]

The final answer is \( \boxed{ AB + CD \leq 2R \sqrt{2(1 - k^2)} } \)