I'll assume by $P^2$ you mean the real projective plane and I'll also assume you know that $\pi_1(P^2)=\mathbb{Z}/2$.
Now, every map $f:(X,x_0) \to (Y,y_0)$ of based topological spaces induces a map between their fundamental groups $f_*:\pi_1(X,x_0) \to \pi_1(Y,y_0)$.
Now, $\pi_1(P^2,x_0)=\mathbb{Z}/2$ (and in fact, we may forget about the base point, since $P^2$ is path connected). Also, $\pi_1(S^1, x_0)=\mathbb{Z}$. So a continous map $f:P^2 \to S^1$ induces, by the above paragraph, a map $\mathbb{Z}/2 \to \mathbb{Z}$. Torsion elements are sent to torsion elements, but $\mathbb{Z}$ is torsion-free, so the only possibility is that $\mathbb{Z}/2$ is mapped to zero. So every map $f:P^2 \to S^1$ induces the trivial map on fundamental groups. This means that $f_*$ is homotopic to the null map. EDIT: See SL2's answer below for the conclusion.
(regarding maps $\mathbb{Z}/2 \to \mathbb{Z}$: What happens to $1 \in \mathbb{Z}/2$? The map must satisfy $f(1+1)=f(1)+f(1)$, but $f(1+1)=f(0)=0$ , so $2f(1)=0$, hence $f(1)=0$.)