Record the player's progress as a path through points $(n, y),$ where $y$ is the number of shots made after $n$ attempts. The condition $a_n \le .4$ caps $y$ at $\lfloor 0.4n \rfloor,$ which for $n = 1, \ldots, 9$ is $0, 0, 1, 1, 2, 2, 2, 3, 3,$ and $a_{10} = .4$ means the path ends at $(10, 4).$

Count the allowed paths by adding, at each point, the counts of its two predecessors (a miss keeps $y,$ a make raises it by $1$). The counts at the maximum allowed heights for $n = 3, \ldots, 9$ come out to $1, 2, 2, 5, 9, 9, 23,$ and the tenth shot must be a make, so $23$ shot sequences qualify. Each consists of $4$ makes and $6$ misses, so the probability is $$23 \left(\tfrac{2}{5}\right)^4 \left(\tfrac{3}{5}\right)^6 = \frac{2^4 \, 3^6 \cdot 23}{5^{10}}.$$

Thus $\{p, q, r, s\} = \{2, 3, 23, 5\}$ and $(a, b, c) = (4, 6, 10),$ giving $(2 + 3 + 23 + 5)(4 + 6 + 10)$ $= 33 \cdot 20$ $= 660.$