Units conversion 1 dB compression

For a load of 50 ohm, can somebody explain to me how if P(1-dB) = -30 dBm, this is equivalent to 10mV(peak)?

It's simple conversion: just remember P[W]=U(max)^2/(2*R)

Convert first dBm to watts.

P(Watts)=10^[P(dBm)/10-3]

so a power of -30 dBm in watt is:

P=10^(-6) watts (that is 1 microwatt)

but since P = V^2/R we can calculate the voltage as

V = sqrt(P*R)

numerically

V=sqrt[10^(-6)*50]=7.07 mV

the above calculation returns a RMS value so we need to multiply by sqrt(2) to have the peak value:

Vpeak=7.07*sqrt(2) = 10 mVpeak

This calculation always apply to convert from dBm to volt (given the load), not only to P1dB measurements

Mhm, now you took him the chance to find it out on himself

Fantastic, thanks for your detailed answer guys :)