Answer :
You made a mistake with the probability [tex]p_{j}[/tex], which should be [tex]p_{i}[/tex] in the last expression, so to be clear I will state the expression again.
So we want to solve the following:
Conditioned on this event, show that the probability that her paper is in drawer [tex]j[/tex], is given by:
(1) [tex]\frac{p_{j} }{1-d_{i}p_{i} } , if j \neq i,[/tex] and
(2) [tex]\frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} } , if j = i.[/tex]
so we can say:
[tex]A[/tex] is the event that you search drawer [tex]i[/tex] and find nothing,
[tex]B[/tex] is the event that you search drawer [tex]i[/tex] and find the paper,
[tex]C_{k}[/tex] is the event that the paper is in drawer [tex]k, k = 1, ..., n.[/tex]
this gives us:
[tex]P(B) = P(B \cap C_{i} ) = P(C_{i})P(B | C_{i} ) = d_{i} p_{i}[/tex]
[tex]P(A) = 1 - P(B) = 1 - d_{i} p_{i}[/tex]
Solution to Part (1):
if [tex]j \neq i[/tex], then [tex]P(A \cap C_{j} ) = P(C_{j} )[/tex],
this means that
[tex]P(C_{j} |A) = \frac{P(A \cap C_{j})}{P(A)} = \frac{P(C_{j} )}{P(A)} = \frac{p_{j} }{1-d_{i}p_{i} }[/tex]
as needed so part one is solved.
Solution to Part(2):
so we have now that if [tex]j[/tex] = [tex]i[/tex], we get that:
[tex]P(C_{j}|A ) = \frac{P(A \cap C_{j})}{P(A)}[/tex]
remember that:
[tex]P(A|C_{j} ) = \frac{P(A \cap C_{j})}{P(C_{j})}[/tex]
this implies that:
[tex]P(A \cap C_{j}) = P(C_{j}) \cdot P(A|C_{j}) = p_{i} (1-d_{i} )[/tex]
so we just need to combine the above relations to get:
[tex]P(C_{j}|A) = \frac{p_{i} (1-d_{i} )}{1-d_{i}p_{i} }[/tex]
as needed so part two is solved.