Understanding digital MEMS microphone sensitivity and obtained bits for a given input SPL
The mems microphone produces a digital signal and when the incident RMS pressure is 94 dB SPL, the digital signal has an RMS level of -26 dBFS. If the positive full scale is (say) \$2^{15}\$ or 32,768 then the digital RMS level is 26 dB down on this at 1,642. This will be at 1 kHz.
1,642 is an RMS number so the full peak-to peak value (for a sinewave) is 4644 LSbs. This cannot be represented by signed number less than 12 bits (13 bits including sign).
83 dB SPL is 11 dB down on 94 dB SPL hence the RMS digital signal level will be 11 dB down on 1,642 i.e. 463. Because you are using a sine wave this translates to a peak-to-peak reading in digital counts of 2.828x higher or 1309 counts p-p.
0.5 watts is 5 times more power than 0.1 watts or an increase of 3.16 dB and, accordingly, the SPL will be 3.16 dB higher and ditto the digital signal produced.
I have no clue on this because I don't know the frequency responses of either speaker or microphone nor the ambient acoustics.
This is tricky to predict because if the equivalent noise is 100 counts RMS (gaussian) then there will be a 99.9% probability it will be within a digital level of 660 digital counts p-p (see below). What is an acceptable level only you can decide.


Gaussian noise and probability: -