1. Interpret the given sum

We need to evaluate

$${}^1P_1 + 2\,{}^2P_2 + 3\,{}^3P_3 + \cdots + 15\,{}^{15}P_{15}.$$

Recall that

$${}^nP_n = n!$$

so the sum becomes

$$1\cdot 1! + 2\cdot 2! + 3\cdot 3! + \cdots + 15\cdot 15!.$$

2. Use a standard simplification

Note that

$$k\cdot k! = (k+1-1)k! = (k+1)! - k!.$$

Therefore,

$$\sum_{k=1}^{15} k\cdot k! = \sum_{k=1}^{15} \big((k+1)! - k!\big).$$

Write out a few terms:

$$(2!-1!) + (3!-2!) + (4!-3!) + \cdots + (16!-15!).$$

This is a telescoping sum, so everything cancels except:

$$16! - 1! = 16! - 1.$$

Thus,

$${}^1P_1 + 2\,{}^2P_2 + \cdots + 15\,{}^{15}P_{15} = 16! - 1.$$

3. Match with the given form

Given

$$16! - 1 = {}^qP_r - s, \qquad 0\le s \le 1.$$

Since $s$ is between $0$ and $1$ and is intended to make the expression match exactly, take

$${}^qP_r = 16!, \qquad s=1.$$

Now,

$$16! = {}^{16}P_{16}.$$

So we can identify

$$q=16,\quad r=16,\quad s=1.$$

4. Compute the required combination

We need

$${}^{q+s}C_{r-s} = {}^{16+1}C_{16-1} = {}^{17}C_{15}.$$

Using symmetry,

$${}^{17}C_{15} = {}^{17}C_2 = \frac{17\cdot 16}{2} = 136.$$

5. Final answer

$$\boxed{136}$$