Let $x$ be a signal and $n$ be an independent noise. How one can show $P(ax+n|x)=P(n)$? Thanks.
Well, let $y=ax+n$, so we have $n=y-ax$.
Now if the probability density function (PDF) of $n$ for example be Gaussian $\mathcal N(\mu_n, \sigma_n)$, i.e. $n\sim \mathcal N(\mu_n, \sigma_n)$), then $(y-ax)\sim \mathcal N(\mu_n,\sigma_n)$.
Let $X$ and $N$ be discrete random variables and, assuming that $P(X=x)>0,$ consider the following conditional probability: $$P(aX+N=s|X=x).$$ Given the event that $X=x$, if $N$ has a value $n=s-ax$, which it takes with a positive probability, then $$P(aX+N=s|X=x)=\frac{P(X=x,ax+N=s)}{P(X=x)}=\frac{P(X=x)P(ax+N=s)}{P(X=x)}$$ $$=P(ax+N=s)=P(N=s-ax)=P(N=n).$$ Given the event that $X=x$ if $N$ doesn't have a suitable value $n$, which it takes with positive probability then $$P(aX+N|X=x)=P(N=n)=0.$$
I am not sure if the above argumentation corresponds to the script below $$P(ax+n|x)=P(n).$$