A different route: check each option by verifying its five (or four) product terms cell by cell instead of redrawing the whole map.
Step 1: List the target minterms. \(F = \sum m(1,2,3,4,5,7,10,12,13,14)\), so the missing (0) minterms are 0, 6, 8, 9, 11, 15. Any candidate expression is correct only if it produces exactly the 10 minterms above and never touches a 0-minterm.
Step 2: Decode each product term into the minterms it turns on. \(P'S\) turns on all minterms with \(P=0, S=1\), that is 1, 3, 5, 7. \(QR'\) turns on all minterms with \(Q=1, R=0\), that is 4, 5, 12, 13. These two terms together give 1, 3, 4, 5, 7, 12, 13 -- still missing 2, 10, 14, and every option must supply these three from its remaining two terms.
Step 3: Test the remaining terms of each option against 2, 10, 14 only.
Option 1 remaining terms \(P'Q'R, Q'RS'\): \(P'Q'R\) gives minterms 2, 3; \(Q'RS'\) gives minterms 2, 10. Union = {2,3,10}. Minterm 14 never appears, so Option 1 fails.
Option 2 remaining terms \(P'Q'R, PRS'\): \(P'Q'R\) gives 2, 3; \(PRS'\) gives 10, 14. Union with earlier set = {1,2,3,4,5,7,10,12,13,14}, exactly matching F. Option 2 works.
Option 3 remaining terms \(PQS', PRS'\): \(PQS'\) gives 12, 14; \(PRS'\) gives 10, 14. Union = {10,12,14}, minterm 2 never appears, so Option 3 fails.
Option 4 remaining terms \(PQS', Q'RS'\): \(PQS'\) gives 12, 14; \(Q'RS'\) gives 2, 10. Union with earlier set = {1,2,3,4,5,7,10,12,13,14}, exactly matching F. Option 4 works.
Step 4: Confirm no 0-minterm sneaks in. None of \(P'Q'R\), \(PRS'\), \(PQS'\), \(Q'RS'\) produces 0, 6, 8, 9, 11 or 15 (each can be checked directly from its defining literals), so both Option 2 and Option 4 stay consistent with F everywhere.
Final answer: options 2 and 4 are the correct minimal sum-of-products expressions.