If we define a polyhedron $$ D:=\{x \in \mathbb{R}^{n}\mid Bx=d,x \geq 0\}, $$ where $B \in \mathbb{R}^{m \times n}$ and $d \in \mathbb{R}^{m}$, the projection problem associates to $D$ is $$ \min \ \frac{1}{2}\|x-c\|^{2}, \ \text{subject to } Bx=d,x \geq 0. $$ The equivalent problem is $$ \min \ \frac{1}{2}\|x-c\|^{2}+\delta_{C}(x), \ \text{subject to } Bx=d, $$ where $C$ is defined by $C:=\{x \in \mathbb{R}^{n}\mid x \geq 0\}$ and \delta_{C}(x) is the indicator function, which is defined as if $x \in C$ equals to 0 and $+\infty$ otherwise.
The Lagrangian function is $$ l(x,y):= \frac{1}{2}\|x-c\|^{2}+\delta_{C}(x)-\left<y,Bx-d\right>. $$ Therefore, the dual problem is $$ \max_{y}\min_{x}\frac{1}{2}\|x-c\|^{2}+\delta_{C}(x)-\left<y,Bx-d\right>. $$ Consider the minimization problem, a equivalent reformulation is $$ \min_{x} \ \frac{1}{2}\|x-c-B^{\top}y\|^{2}+\delta_{C}(x)+\left<y,d\right>-\left<B^{\top}y,c\right>-\frac{1}{2}\|B^{\top}y\|^{2} $$ Then, we can conclude that $x=\Pi_{C}(c+B^{\top}y)$ with \Pi_{C} a projection onto $C$. Thus, the dual problm is $$ \max_{y} \frac{1}{2}\|\Pi_{C}(c+B^{\top}y)\|^{2}+\frac{1}{2}\|c\|^{2}-\left<\Pi_{C}(c+B^{\top}y),c+B^{\top}y\right>+\left<d,y\right>. $$ However, the answer is $$ \max_{y} -\frac{1}{2}\|\Pi_{C}(c+B^{\top}y)\|^{2}+\frac{1}{2}\|c\|^{2}+\left<d,y\right>. $$ I do not know where I make some mistakes.
Both expressions coincide.
Hint: Try to simplify $$ \langle \Pi_C(z), z \rangle. $$