Following up on Dai's answer, one can go a step further since $P U(H)$ is obviously a group. So if we can find a contractible space on which it acts freely, the quotient will be the next level up (namely, a $K(\mathbb{Z},3)$.

Such a space can be constructed as follows: take our favourite (separable, though that's not necessary) Hilbert space, $H$, and consider $HS(H)$, the space of Hilbert-Schmidt operators on $H$. This is isomorphic to the Hilbert tensor product $H^* \widehat{\otimes} H$ so is a Hilbert space. Its unitary group is thus contractible. The group $U(H)$ acts on $HS(H)$ by conjugation, and once we divide out by the centre this becomes free. Thus $P U(H)$ acts on $U(HS(H))$ freely and so the quotient is a $K(\mathbb{Z},3)$.

However, as $P U(H)$ does not act centrally on $U(HS(H))$, the iteration stops here.

Following up on Dai's answer, one can go a step further since $P U(H)$ is obviously a group. So if we can find a contractible space on which it acts freely, the quotient will be the next level up (namely, a $K(\mathbb{Z},3)$).

Such a space can be constructed as follows: take our favourite (separable, though that's not necessary) Hilbert space, $H$, and consider $\operatorname{HS}(H)$, the space of Hilbert–Schmidt operators on $H$. This is isomorphic to the Hilbert tensor product $H^* \mathbin{\widehat{\otimes}} H$ so is a Hilbert space. Its unitary group is thus contractible. The group $U(H)$ acts on $\operatorname{HS}(H)$ by conjugation, and once we divide out by the centre this becomes free. Thus $P U(H)$ acts on $U(\operatorname{HS}(H))$ freely and so the quotient is a $K(\mathbb{Z},3)$.

However, as $P U(H)$ does not act centrally on $U(\operatorname{HS}(H))$, the iteration stops here.