Factor p(x)=3x+2x2-x3=x(3+2x-x2)=x(3-x)(1+x). Now replace x with a-3:
p(a-3)=(a-3)(3-a+3)(1+a-3)=(a-3)(6-a)(a-2)=
(a-3)(6a-12-a2+2a)=(a-3)(8a-12-a2)=a(8a-12-a2)-3(8a-12-a2)=
8a2-12a-a3-24a+36+3a2=-a3+11a2-36a+36.
So p(a-3)=-a3+11a2-36a+36.
QUICK CHECK: When x=0, p(0)=0, so when a-3=0, that is, a=3 we should get 0 when we plug a=3 into p(a-3):
p(3-3)=p(0)=-27+99-108+36=135-135=0✔️ This proves the answer (in bold).
NOTE: You don't have to factor p(x); you can simply replace x with a-3 in 3x+2x2-x3, but you would end up with more terms to add or subtract and you could easily make a mistake.