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Correct Option is D, S, B

From the given statements, A has 3 pens more than F and sits immediate left to F, who faces the one who has 10 pens more than F. A does not face inside. From these conditions we get 2 possibilities i.e. Case 1 and Case 2. P has 32 pens and sits 2nd to the left and 3rd to the left of the one who has 25 pens and 49 pens respectively.
S has pens in square value of odd number which is more than five and sits 3rd to the left of D, who has 1 pen more than A. From this condition only one option is left i.e., S has 49 pens.R has 10 pens more than A and faces to D. From this condition Case 2 is ruled out now. C sits 2nd to the right of G but not an immediate neighbour of D and C has 1 pen more than G. Q is neither an immediate neighbour of S nor D.H sits 2nd to the right of B and has less pens than Q. Difference between the number of pens of H and S are not more than 8. From these conditions we have only option left that H has 42 pens. Q has more pens than E. So, the final arrangement is-

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