$\rho$-shift in parabolically induced representations In order to define the principal series representations one takes an irreducible, unitary representation $\sigma$ of $M$ (here $G$ is a semisimple Lie group with Iwasawa decomposition $G=KAN$ and $M$ is the centralizer of $A$ in $K$) and a character $\nu\in\mathfrak{a'}_\mathbb{C}$ to obtain an $MA$-representation. One extends trivially to get an $MAN$-module. Now one induces this representation to $G$. However the induced space consists  of functions $f$ that satisfy $f(kman)=e^{-(\nu+\rho)(\log a)}\sigma(m)^{-1}f(k)$.
Why is there a $\rho$-shift and for which reason does one need it?
 A: Let $G$ be a Lie group and $H$ a closed subgroup of $G$. In general, the unitarity of a representation of $H$ is not preserved when inducing unless there is a choice of invariant positive density (positive Haar measure) on $G/H$. To avoid this problem, we tensor the original representation with the one-dimensional representation given by the square root of the modular function on $H$, when constructing the induced representation. In other words, if $(\xi, V_\xi)$ is a continuous finite dimensional Hilbert representation of $H$ and if we denote by $\Delta$ the modular function on $H$, then the tensor product representation $(\xi\otimes\Delta^{1/2}, V_\xi\otimes\mathbb{C})$ can be isometrically realised in $V_\xi$ by the representation
$$(\xi\otimes\Delta^{1/2})(h)v = \Delta(h)^{1/2}\xi(h)v\quad \text{for}\, v\in V_\xi, h\in H$$
(Here I have slightly abused the notation). Now we induced from this last representation and we get ${\rm Ind}_H^G(\xi) = {\rm ind}_H^G(\xi\otimes\Delta^{1/2})$ (Note the capital letters). Then ${\rm Ind}_H^G$ is the normalised induced representation.
Answering your question, the principal series representation is $\textbf{normalised}$ parabolic induction with respect to certain characters. Thus, with your $MAN$-module $V_\sigma$ given by $$(\sigma\otimes\nu)(man)v = e^{\nu({\rm log}\, a)}\sigma(m)v$$
you still have to tensor it with the square root of the modular function of $P$. We claim that $\Delta_P(man) = e^{2\rho_P({\rm log}\, a)}$. This means that 
$$((\sigma\otimes\nu)\otimes \Delta_P^{1/2})(man) = e^{(\nu + \rho)({\rm log}\, a)}\sigma(m) = (\sigma\otimes(\nu + \rho))(man)\quad\text{for}\, man\in MAN$$
Thus, we observe that normalising comes just about as shifting by $\rho$.
$\textbf{Claim}$: $\Delta_P(man) = e^{2\rho_P({\rm log}\,a)}$.
$\textit{Proof.}$ First of all, since $M$ is compact, $\Delta_P(M) = 1$. Then
$$\Delta_P(an) = \frac{|\det {\rm Ad}(an)|_{\mathfrak{m}\oplus\mathfrak{a}\oplus\mathfrak{n}}|}{|\det {\rm Ad}(an)|} = |\det{\rm Ad}(an)|_{\overline{\mathfrak{n}}}| = |\det{\rm Ad}(a)|_{\overline{\mathfrak{n}}}|^{-1}|\det{\rm Ad}(n)|_{\overline{\mathfrak{n}}}|^{-1}$$
One can also check that ${\rm Ad}(n)$ is a nilpotent map for every $n\in N$, hence the second term of the product is equal to 1. Then we have that
$$\Delta_P(man) = |\det{\rm Ad}(a)|_{\overline{\mathfrak{n}}}|^{-1} = e^{-{\rm tr}(ad({\rm log}\, a)|_{\overline{n}})} = e^{2\rho_P({\rm log}\, a)} $$
