Suppose $P_L : H \rightarrow H$ is orthogonal projection of closed subspace L. Then, I would like to prove that $P_L = P_L^{*}$ i.e it is the same as the adjoint.
Suppose $l \in L$, then consider $l - P^{*}_L(l)$. Then we have that $(l,l - P_L^{*}(l)) = (l,l) - (l,l) = 0$ by main property of the dual.
I am having troubles figuring out why $P_L^*(l) = l$.
I assume that the problem refers to $ H $ as a Hilbert space.
By definition of adjoint map $P^{*}_{L}$ is the only map the satisfies $$\left\langle P_{L}(x),y\right\rangle = \left\langle x, P^{*}_{L}(y)\right\rangle $$ for all $x,y\in H$.
For other hand, given $x,y\in H$, since $L$ is closed subspace we have $H=L \oplus L^{\bot}$, then $x=P_{L}(x)+(x-P_{L}(x))$ and $y=P_{L}(y)+(y-P_{L}(y))$ where $x-P_{L}(x)\in L^{\bot}$ and $y-P_{L}(y)\in L^{\bot}$, that is $$\left\langle P_{L}(x),y-P_{L}(y)\right\rangle=0 \quad \mbox{and}\quad \left\langle x-P_{L}(x),P_{L}(y)\right\rangle=0. $$
Therefore $$\begin{array}{rcl} \left\langle P_{L}(x),y\right\rangle &=& \left\langle P_{L}(x),P_{L}(y)+(y-P_{L}(y))\right\rangle\\ &=& \left\langle P_{L}(x),P_{L}(y)\right\rangle +\left\langle P_{L}(x),y-P_{L}(y)\right\rangle\\ &=& \left\langle P_{L}(x),P_{L}(y)\right\rangle \\ &=& \left\langle P_{L}(x),P_{L}(y)\right\rangle + \left\langle x-P_{L}(x),P_{L}(y)\right\rangle \\ &=& \left\langle P_{L}(x)+(x-P_{L}(x)),P_{L}(y)\right\rangle\\ &=& \left\langle x, P_{L}(y)\right\rangle. \end{array}$$
By the uniqueness of $P^{*}_{L}$ we have $P_{L}=P^{*}_{L}$.