Homotopy group with coefficients

Remarks

For $n \ge 3$ the space $P^n(k)$ is the suspension of $P^{n-1}(k)$, so $\pi_n(X; \Z/k)$ is a group. For $n \ge 4$ it is a double suspension, so the group is abelian. For $n = 2$ the set $\pi_2(X; \Z/k)$ is in general only a pointed set.

For $n \ge 3$ the cofibre sequence $S^{n-1} \xrightarrow{k} S^{n-1} \to P^n(k)$ gives a short exact sequence $$0 \to \pi_n(X) \otimes \Z/k \to \pi_n(X; \Z/k) \to \set{x \in \pi_{n-1}(X) : kx = 0} \to 0,$$ where $\pi_n(X)$ is the ordinary homotopy group.

Example: for $X = S^n$ with $n \ge 3$, $\pi_n(S^n) \otimes \Z/k \cong \Z/k$ and $\pi_{n-1}(S^n) = 0$, so $\pi_n(S^n; \Z/k) \cong \Z/k$.

Non-example: maps from $S^n$ to $X$ do not give $\pi_n(X; \Z/k)$, since $S^n$ is not a Moore space $M(\Z/k, n-1)$; its homology in degree $n - 1$ is $0$ for $n \ge 2$, not $\Z/k$.