What do you mean?

Perhaps this?

f[p0_, p1_] := 
    Module[{},
   Subscript[p, 0] = p0;
    Subscript[p, 1] = p1; 
      1/(2 bb n (Subscript[p, 0] + (-1 + n) Subscript[p, 1])) 
    \!
\*SubsuperscriptBox[\(p\), \(0\), \(-n\)] (-2 (-100 + d2) 
       \!
\*SubsuperscriptBox[\(p\), \(0\), \(2\)] + 
            2 bb (d2 + 100 (-1 + n)) 
       \!
\*SubsuperscriptBox[\(p\), \(0\), \(2 + n\)] - bb (-1 + n) n 
       \!
\*SubsuperscriptBox[\(p\), \(0\), \(n\)] Subscript[p, 
               1] (200 + (-100 + 2 d1 - d2) Subscript[p, 1]) + 2 bb n 
       \!
\*SubsuperscriptBox[\(p\), \(0\), \(1 + 
         n\)] (-100 + (-d1 + d2 + 100 (-1 + n)) Subscript[p, 1]))];


f[a, b]

Out[2]= (1/(2 bb (a + b (-1 + n)) n))a^-n (-2 a^2 (-100 + d2) + 
   2 a^(2 + n) bb (d2 + 100 (-1 + n)) + 
   2 a^(1 + n) bb (-100 + b (-d1 + d2 + 100 (-1 + n))) n - 
   a^n b bb (200 + b (-100 + 2 d1 - d2)) (-1 + n) n)