(a) Consider the 3 Es as a single unit. This gives us the units: S, H, O, P, K, (EEE), P, R. There are 8 units in total. The number of ways to arrange these 8 units is \(\frac{8!}{2!}\) because the Ps are repeated. Therefore, the number of ways is \(\frac{8!}{2!} = 20160\).

(b) The total number of ways to arrange the letters of SHOPKEEPER is \(\frac{10!}{2!3!} = 302400\). If the Ps are together, treat them as a single unit: S, H, O, (PP), K, E, E, E, R. This gives us 9 units. The number of ways to arrange these 9 units is \(\frac{9!}{3!} = 60480\). Therefore, the number of ways the Ps are not together is \(302400 - 60480 = 241920\).

(c) The number of ways to have Es at the beginning and end is the number of ways to arrange the remaining 8 letters: S, H, O, P, K, E, P, R. This is \(\frac{8!}{2!} = 20160\). The total number of arrangements is \(\frac{10!}{2!3!} = 302400\). Therefore, the probability is \(\frac{20160}{302400} = \frac{1}{15}\).