Prove that the projection operator $\mathbb P_+\equiv|+z\rangle\!\langle +z|$ is Hermitian

Any projection operator can be written in the form $$ P = \sum_{j = 1}^r |\psi_j \rangle \langle \psi_j | $$ Where $\psi_1,\dots,\psi_n$ is an orthonormal basis of our Hilbert space. Given $\psi = c_1\psi_1 + \cdots + c_n \psi_n$, we calculate $$ \langle\psi| P = \langle \psi | \left(\sum_{j = 1}^r |\psi_j \rangle \langle \psi_j | \right) = \sum_{j = 1}^r \langle \psi \mid \psi_j \rangle \langle \psi_j | =\\ \sum_{j = 1}^r \langle \psi_j \mid \psi \rangle^* \langle \psi_j | = \sum_{j = 1}^r c_j^* \langle \psi_j | $$ This is the bra corresponding to the ket $P |\psi \rangle = \sum_{j=1}^r c_j | \psi_j \rangle$. So, $P$ is self-adjoint.

I assume from the comments that $P$ is a rank-1 projection, and so it is of the form $$P=|\psi\rangle\langle\psi|$$ for some vector $|\psi\rangle$ in the Hilbert space. Observe that, in a sense, $|\psi\rangle^*=\langle\psi|$ in Dirac notation, whence $$P^* = (\langle\psi|)^*(|\psi\rangle)^*,$$ and since the $*$ is involutive one has $$P^*=|\psi\rangle\langle\psi| = P.$$

The "idempotency" $P^2 = P$ comes from a direct computation $$P^2 = |\psi\rangle\langle\psi|\psi\rangle\langle\psi|=|\psi\rangle\langle\psi|=P$$ since $|\psi\rangle$ is assumed to be a vector of norm one, so that $\Vert\psi\Vert^2 = \langle\psi|\psi\rangle=1$.