About constants in fermionic path integral in Peskin and Schroeder

If you mean that the factor of $i$ is inconsistent, note that $$ det (i M) = i^d det M $$ where $d = dim M$ is the number of degrees of freedom that we integrate over. Hence, $i^d$ is a constant.

With path integrals, $d = infty$ and it is not clear how to make sense of $i^d$. This problem can be traced back to the definition of the path integral.

In physics, we usually treat path integrals as abstract symbols that conveniently encode the properties that we expect the actual mathematically well-defined quantities (correlation functions) to have. The definitions of correlation functions naively don't depend on a normalization constant: $$ left< Omega right> = frac{int D phi e^{i S[phi]} Omega[phi]}{int D phi e^{i S[phi]}} = frac{N int D phi e^{i S[phi]} Omega[phi]}{N int D phi e^{i S[phi]}}. $$

Hence, an arbitrary constant factor can be absorbed into the definition of the path integral measure.

Ofcourse, one has to always keep in mind that the value of the path integral is defined up to an arbitrary normalization constant. For example, saying that $det M = 42$ in your example wouldn't mean anything physically relevant – I can always make it $det M = 43$ by absorbing $N = 43/42$ in the path integral measure.

What is physically relevant is the dependence of $det M(varepsilon)$ on $varepsilon$, where $varepsilon$ is some parameter that is meaningful for your problem (e.g. in QFT path integrals this is usually the "source field"). One can compare e.g. $$ frac{det M(varepsilon_1)}{det M(varepsilon_2)} = frac{N det M(varepsilon_1)}{N det M(varepsilon_2)} $$ which is independent of the choice of $N$.