just find $ a $ in terms of $s$ and $t$...and then subsitute .

Put b-st in the last equation which means we have to prove b-st+a+1 =0. Subtract P(t) and P(s) to get a+1 = -s-t. All that remains to prove is that b-st-s-t =0. For this add P(s) and P(t) and substitue a = -s-t-1. You get the result

Yeah, this one was easy.

see attachment for my solution

Attachments:

troll question my solution = by $p(s)=s$ and $p(t)=t$ we easily get $a+s+t=-1$ now we prove that $p(s).p(t)-p(b-st)=0$ which on expanding and using $a+s+t=-1$ cancels to $0$. so we are done.

@aditya21 $P(s) \neq P(t)$

kgdpsdurg wrote: @aditya21 $P(s) \neq P(t)$ corrected