It is true that $\mathfrak p$ is always an irreducible $\mathfrak k$-module. Be aware though that the complexification $\mathfrak p_{\mathbb C}$ might be reducible as an $\mathfrak k_{\mathbb C}$-module. This happens precisely in the case that $\mathfrak g$ is of Hermitian type.

It is true that $\mathfrak p$ is always an irreducible $\mathfrak k$-module. Be aware though that the complexification $\mathfrak p_{\mathbb C}$ might be reducible as an $\mathfrak k_{\mathbb C}$-module. This happens precisely in the case that $\mathfrak g$ is of Hermitian type.

Edit: If $\tau$ is not a Cartan involution then $\mathfrak g^{-\tau}$ may be a reducible $\mathfrak g^\tau$-module even over $\mathbb R$. The criterion is whether the center of $\mathfrak g^\tau$ is split or not. A typical example is $\mathfrak g=\mathfrak{sl}(2,\mathbb R)$ and $\tau$ is conjugation by $diag(1,-1)$.

It is true that $\mathfrak p$ is always an irreducible $\mathfrak k$-module. By aware though that the complexification $\mathfrak p_{\mathbb C}$ might be reducible as an $\mathfrak k_{\mathbb C}$-module. This happens precisely in the case that $\mathfrak g$ is of Hermitian type.

It is true that $\mathfrak p$ is always an irreducible $\mathfrak k$-module. Be aware though that the complexification $\mathfrak p_{\mathbb C}$ might be reducible as an $\mathfrak k_{\mathbb C}$-module. This happens precisely in the case that $\mathfrak g$ is of Hermitian type.