(a) \(9!/(2!2!)=90720\).

(b) Only PP and SS can give equal end letters. Each gives \(7!/2!\) arrangements. \[90720-2\frac{7!}{2!}=85680.\]

(c) If the Ps lie between the Ss, the block SPPS and five distinct letters give \(6!=720\) arrangements. Otherwise form an \(S\_\_S\) block by choosing and ordering two of the five single letters; this block, PP and the three remaining letters give \[5\cdot4\cdot5!=2400.\] The total is \(720+2400=3120\).

(d) Choose the group of 5 from the 9 individual tiles. If it contains PP and no S, choose 3 of the 5 single letters; if it contains SS and no P, the same applies. \[P=\frac{2\binom53}{\binom95}=\frac{20}{126}=\frac{10}{63}=0.159.\]