You can just replace $\sigma$ in
$$ P(x) = \frac{1}{{\sigma \sqrt {2\pi } }} \exp \left\{{{{ - \left( {x - \mu } \right)^2 } \mathord{\left/ {\vphantom {{ - \left( {x - \mu } \right)^2 } {2\sigma ^2 }}} \right.} {2\sigma ^2 }}} \right\}$$
with $\exp{\ln{\sigma}}$, thereby yielding the following
$$ P(x) = \frac{1}{{\exp{\ln{\sigma}} \sqrt {2\pi } }} \exp \left\{{{{ - \left( {x - \mu } \right)^2 } \mathord{\left/ {\vphantom {{ - \left( {x - \mu } \right)^2 } {2\sigma ^2 }}} \right.} {2(\exp{\ln{\sigma}})^2 }}} \right\}$$
which can be further rewritten as follows
$$ P(x) = \frac{1}{{\sqrt {2\pi } }} \exp \left\{{{{ - \left( {x - \mu } \right)^2 } \ln{\sigma}\mathord{\left/ {\vphantom {{ - \left( {x - \mu } \right)^2 } {2\sigma ^2 }}} \right.} {2\exp{(2 \ln{\sigma})} }}} \right\}.$$
Since $\sigma$ here is a fixed parameter rather than a random variable, no further details are necessary. However, should $\sigma$ be a random variable with probability $P(\sigma)$, and should you want to use $\ln \sigma$ there, namely $P(\ln \sigma)$, then you should change $P(\sigma)$ accordingly.
Is that what you want? Does this helps? If not, could you reformulate the question? Thanks.
UPDATE:
It seems to me that what they are doing in that post is analogous to the following $$\int{f(x) P(x|\sigma) P(\sigma) d\sigma dx}$$ and that they tried to compute this but directly replacing $\sigma$ with $\ln{\sigma}$. In order to obtain the same results, the integral must be expressed as $$\int{f(x) P(x|\ln{\sigma}) P(\ln{\sigma}) \frac{1}{\sigma} d\sigma dx} $$, that is, by taking into account that changing $\sigma$ into $\ln\sigma$ also changes the differential within the integral. The result would be perhaps clearer is we write $\sigma=\ln{u}$. In that case, the integral would read $$\int{f(x) P(x|\ln{u}) P(\ln{u}) \frac{1}{u} du dx}. $$ I may be wrong about what they tried to do in that post, but this may hopefully help you anyway.
