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This seems to be a typo. Let me explain it again.

Given:
(1) Transmit power of AP is 100 mW, i.e. Pap = 100 mW
(2) Received power of laptop1 is 10 mW, i.e. P1 = 10 mW
(3) Received power of laptop2 is 1 mW, i.e. P2 = 1 mW

(1) The power difference (dB) between AP and laptop1 will be:
y (dB) = 10 x log(Pap / P1) = 10 x log(100 mW / 10 mW) = 10 x log(10/1) = 10 x 1 = 10 dB [Formula 1]

(2) The power difference (dB) between AP and laptop2 will be:
y (dB) = 10 x log(Pap / P2) = 10 x log(100 mW / 1 mW) = 10 x log(100/1) = 10 x 2 = 20 dB [Formula 2]

In the study guide, unfortunately, while calculating the power difference between AP and laptop2, Formula 1 was mistakenly used instead of Formula 2, but the power values of AP and laptop2 were correct. So you see the following:

y = 10 x log(Pap / [b]P1[/b]) = 10 x log (100/1) = 10 x 2 = 20 dB

Note the bold [b]P1[/b] above. It should be [b]P2[/b] since we are calculating the decibels from AP to laotop2.