Applying periodogram-based PSD
It's exactly the same method, only the second is converted to (power) dB ($\text{dB}:\,x\mapsto 10\log_{10}x$):

Let's convert your first formula to dB:

$$\begin{align}
P_{1,\text{dB}} &amp;= 10\log_{10}P_1\\
&amp;=10\log_{10}\left(\frac{\Delta t}{N}\lvert\text{FFT}\rvert^2\right)\\
&amp;=10\log_{10}\left(\frac{\lvert\text{FFT}\rvert^2}{f_sN}\right)\\
&amp;=10\log_{10}\left(\frac{\lvert\text{FFT}\rvert^2}{\sqrt{f_sN}^2}\right)\\
&amp;=10\log_{10}\left(\left(\frac{\lvert\text{FFT}\rvert}{\sqrt{f_sN}}\right)^2\right)\\
&amp;=10\cdot2\cdot\log_{10}\left(\frac{\lvert\text{FFT}\rvert}{\sqrt{f_sN}}\right)\\
&amp;=20\log_{10}\left(\frac{\lvert\text{FFT}\rvert}{\sqrt{f_sN}}\right)\\
\end{align}$$

The $+30$ just happen because your seconds source is considering $\text{dBm}$ (decibel of milliwatt) instead of $\text{dBW}$ (decibel of watt), assuming your input values are in Volt over a impedance of 1Ω or so. 

Note: you nowhere in your question stated the physical units of what your input data is; be very careful when "implying" something can be represented as $\text{dBm}$ or $\text{dBW}$. That makes a statement about your signals origin (namely, that your ADC or wherever the signal came from is calibrated, or will be converted to analog with calibrated equipment) if you're dealing with real-world observations/signals. This assumption is usually wrong – if you just got a good microphone and a good sound card (or a good receiver and a good SDR device) your digital amplitudes will be proportional, but not identical, to the voltages observed, or even more significantly, to the sound intensity (or EM field strength) that caused the electrical signal in the first place.