Can PSNR > 48 used to validate the signals between two audio files?
If the reconstructed audio signal is exactly same as original signal then MSE =0. And if Max pixel value is 255 (8-bit representation), then the value of PSNR = 20*log(255) = 48dB


That statement is plain wrong. If the two signals are identical, then MSE = 0, then division by zero is undefinied and you don't get a PSNR at all, but "infinity".


  Since the PSNR of the songs are more than 48, can I conclude that the decoded audio signal in the wav file is valid?


PSNR is really not a very useful metric; you can get these values with a codec that simply broken, or one that sounds great.

Let's throw a bit of math in here: let's say we map our 8 bit to - to +1. Our original audio signal $S$ is well-scaled and its amplitude is pretty normally distributed¹, centered around 0, and has a variance $\sigma^2=\frac14$, so that clipping very rarely happens, i.e. $S\sim\mathcal N(0;\frac14)$.

Then, what's the PSNR of a codec that simply always produces "0"?

\begin{align}
\text{PSNR} &amp;= 20\log_{10}\left(\frac{\text{Maxval}}{\sqrt{\text{MSE}}}\right)\\
&amp;=20\log_{10}\left(\frac{\text{1}}{\sqrt{\text{MSE}}}\right)\\
&amp;=10\log_{10}\left(\frac{\text{1}}{\sqrt{\text{MSE}}}\right)^2\\
&amp;=10\log_{10}\left(\frac{\text{1}}{\text{MSE}}\right)\\
\end{align}

So, what's our MSE? It is 

\begin{align}
\mathbb E\left((0-S)^2\right)&amp;= \mathbb E\left( S^2\right)\\
&amp;S \text{ is zero-mean}\\
&amp;=\text{Var}\{S\}=\sigma^2\\
&amp;=\frac14
\end{align}

Therefore, 

\begin{align}
\text{PSNR} &amp;= 10\log_{10}(4) \\
&amp;\approx 6
\end{align}

So, what this would prove is that your codec is better than just producing the mean value of the information-theoretically worst possible continuous audio source – but that's about it.


¹ that assumption about the amplitude distribution is slightly wrong, but wrong in favor of your codec.