Let $H$ be a separable Hilbert space. Let $p \in B(H)$ be a projection (i.e. $p=p^*=p^2$) and $u \in B(H)$ be a unitary (i.e. $uu^*=u^*u= id$).
Consider the operator $v = pup$. If $u$ commutes with $p$ then $vv^* = v^*v = p$.
What about the converse?
The converse is true.
Proof 1:
Take $q_1 = u^*pu$ and $q_2 = upu^*$ and observe that $q_i=q_i^*=q_i^2$, so that there are projections.
Now, by assumption $pu^*pup = pupu^*p = p$, so $pq_ip = p$. It follows that $p \le q_i$. But \begin{align*} p \le q_1 & \Rightarrow upu^* \le uq_1u^* \\ & \Rightarrow q_2 \le p \end{align*}
It follows that $p=q_2 = upu^*$, which means that $pu = up$, i.e. $u$ commutes with $p$.
Proof 2:
First of all \begin{align*} p&= pu^*up = pu^*(p+id-p)up = pu^*pup + pu^*(id-p)up \\ &=puu^*p = pu(p+id-p)u^*p = pupu^*p + pu(id-p)u^*p. \end{align*}
But, by assumption $pu^*pup = pupu^*p = p$. Then \begin{align*} 0& = pu^*(id-p)up = ((id-p)up)^*((id-p)up) \\ & = pu(id-p)u^*p = (pu(id-p))(pu(id-p))^*. \end{align*} Take $A = (id-p)up$ and $B = pu(id-p)$. Then $A^*A = BB^* = 0$. So $\forall \xi \in H$, \begin{align*} 0& = \langle A^*A \xi, \xi \rangle = \langle A \xi, A \xi \rangle = \Vert A \xi \Vert^2 \\ & = \langle BB^* \xi, \xi \rangle = \langle B^* \xi, B^* \xi \rangle = \Vert B^* \xi \Vert^2. \end{align*} It follows that $0=A=B^*= B$. But $$ u = (p+id-p)u(p+id-p) = pup+ (id-p)u(id-p)+A+B.$$
Conclusion, $u = pup + (id-p)u(id-p)$, i.e. $u$ commutes with the projection $p$.