$\det A = \det (P^{-1} D P)$, pelo Teorema de Binet, temos:$$\det A = \det P \cdot \det D \cdot \frac{1}{\det P} \implies \det A = \det D$$Ademais, $A^2 + A = A\cdot (A+I_3)$, calculemos $A+I_3$, sabendo que $I_3 = P^{-1}P$ :$$A+I_3 = P^{-1} D P + P^{-1}P = P^{-1}\cdot (D+I_3)\cdot P \implies \det(A+I_3) = \det(D+I_3) $$Assim, temos:$$\det(A^2 + A) = \det[D\cdot (D+I_3)] = \det(D) \cdot \det(D + I_3)$$$$\det (D) = 1\cdot 2 \cdot 3 = 6\space \space \space \text{e} \space \space \det (D+I_3) = 2\cdot 3 \cdot 4 = 24$$Portanto, $\boxed{\det(A^2 + A) = 6\cdot 24 = 144}$$$\text{Alternativa } \mathbb{(A)}$$