‘MPQO’ is related to ‘KNSQ’ in a certain way based on the English alphabetical order. In the same way, ‘DGHF’ is related to ‘BEJH’. To which of the following is ‘EINW’ related, following the same logic?

Correct — D.

Rule: the first two letters move back 2, the last two move forward 2.

MPQO → KNSQ: M−2=K, P−2=N, Q+2=S, O+2=Q.
DGHF → BEJH: D−2=B, G−2=E, H+2=J, F+2=H.

Apply the same split to EINW:
E−2=C, I−2=G, N+2=P, W+2=Y.

That gives CGPY, which is option (d).

Why the others are wrong

• (a)
  CHQY — CHQY gets the outer letters right but not the inner pair. H is I moved back only one, and Q is N moved forward three, where the rule needs G and P.
• (b)
  BHQX — None of B, H, Q, X matches the −2, −2, +2, +2 shifts. B is E moved back three and X is W moved forward only one.
• (c)
  BGPX — The middle pair G and P is correct, but the outer letters are not: B is E moved back three, and X is W moved forward only one.

Four-letter analogies are position arithmetic. Convert to alphabet numbers, take the difference letter by letter, and the rule falls out in one pass.

The shape here is a split rule: the first half moves back, the second half moves forward. Check all four positions before committing — a rule that fits the first letter and fails on the third is the usual way these are lost.

Two worked pairs are given rather than one, and the rule has to fit both. MPQO → KNSQ on its own would leave a uniform shift open as a reading.

Key facts

• M−2=K, P−2=N, Q+2=S and O+2=Q, so MPQO gives KNSQ.
• D−2=B, G−2=E, H+2=J and F+2=H, so DGHF gives BEJH.
• E−2=C, I−2=G, N+2=P and W+2=Y, so EINW gives CGPY.

Common traps

• Applying the first half's −2 to all four letters, which gives CGLU and matches no option.
• Losing count near the end of the alphabet, where W+2 must land on Y and not Z.

No directly related past PYQ was found.