The four prime numbers have to fulfill $p > 5$ and $s < p + 10$ and hence they must be among the five consecutive odd numbers $p$, $p + 2$, $p + 4$, $p + 6$ and $p + 8$.
As we have to choose 4 out of the five numbers $p$, $p + 2$, $p + 4$, $p + 6$, $p + 8$, we have to omit exactly one of these numbers. If we omit one of the numbers $p$, $p + 2$, $p + 6$ or $p + 8$, three subsequent odd numbers remain, one of which has to be divisible by 3, which is excluded.
Therefore, we have to omit $p + 4$.
Hence the four prime numbers have to be $p$, $q = p + 2$, $r = p + 6$ and $s = p + 8$.
Exactly one of the five consecutive integers $p$, $p+2$, $p+4$, $p+6$ and $p+8$ is divisible by 5.
By construction, none of the chosen number $p$, $q$, $r$, $s$ can be divisible by 5. This implies that $p+4$ is divisible by 5. So $p+4$ is divisible by 15.
The fact that
$$p + q + r + s = p + (p + 2) + (p + 6) + (p + 8) = 4p + 16 = 4(p + 4)$$
yields that the sum is divisible by 60.