EDIT: I found a brief discussion of this in Husemoller's Fibre Bundles, chapter 16 section 12. Here to compute $\tilde K(\mathbb R P^{2n+1})$ he says to consider the map $ \mathbb R P^{2n+1} = S^{2n+1}/\pm 1\to \mathbb C P^n = S^{2n+1}/U(1). $ Under this map the canonical line bundle over $\mathbb C P^n$ pulls back to the complexification of the canonical line bundle over $\mathbb R P^{2n+1}$. Then he says from looking at the (Atiyah-Hirzebruch) spectral sequence we get that $\tilde K(\mathbb R P^{2n+1}) = \mathbb Z/2^n$. I don't see how looking at the spectral sequence helps (and what we learn from the map from $\tilde K(\mathbb C P^{n}) \to \tilde K(\mathbb R P^{2n+1})$). All I can see is that $\tilde K(\mathbb R P^{2n+1})$ is pure torsion (by the Chern character isomorphism) and that it has order $2^n$ (from the spectral sequence). I can't see why, for example, it isn't just a direct sum of $\mathbb Z/2$'s.
I found a homework assignment online from an old K-theory course and one of the problems says to compute $K(\mathbb R P^n)$ by using a suitable comparison map with $\mathbb C P^k$ and knowledge of $K(\mathbb C P^k)$.
I have attempted this but have not been able to get anywhere. The only map $\mathbb R P^n \to \mathbb C P^n$ I can think of is the one sending the equivalence class of $(x_0,\ldots,x_n) \in \mathbb R P^n$ to its equivalence class in $\mathbb C P^n$. Under this (I think) the tautological line bundle over $\mathbb C P^n$ (which generates $K(\mathbb C P^n)$) gets sent to the complexification of the tautological line bundle over $\mathbb R P^n$. But I really don't see where to go from here; if I had a map going the other way maybe I'd be able to say something but the map I have is neither injective nor surjective. I also can't see how torsion is going to come out of this: $K(\mathbb C P^n)$ is torsionfree but $K(\mathbb R P^n)$ isn't.