Answer
$\hat{b}=A(A^TA)^{-1}A^Tb$
Work Step by Step
Let us suppose $\hat{x}$ satisfies $A\hat{x}=\hat{b}$
Because $\hat{b}$ is the orthogonal projection of b onto Col A, $b-\hat{b}=b-A\hat{x}$ is orthogonal to the columns of A. This means $A^T(b-A\hat{x})=\vec{0}$. This means $A^Tb-A^TA\hat{x}=\vec{0}$. From here, we can see that $A^Tb=A^TA\hat{x}$ and $\hat{x}=(A^TA)^{-1}A^Tb$.
Therefore, $\hat{b}=A(A^TA)^{-1}A^Tb$
