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The matrix $X$ is of rank $m$ almost surely (a.s.). So, a.s. the Moore--Penrose inverse of $X$ is $X^+=X^\top(XX^\top)^{-1}$ and hence $$X^+X=X^\top(XX^\top)^{-1}X.$$ Assuming that $\det\Sigma\ne0$ and letting $$Z:=S^{-1/2}X,$$ where $S:=I_m\otimes\Sigma$ (the Kronecker product), we see that the rows $z_i$ of the matrix $Z$ are iid $N(0,I_n)$, and $$X^+X=X^\top(XX^\top)^{-1}=Z^\top(ZZ^\top)^{-1}Z=Z^+Z.$$ So, the problem reduces to the case when $\Sigma=I_n$, which you said is easy.