a) For given $i, j \in \{1, 2, \dots, n\}$, by computing the expansion along the row $i$, we obtain $\det(A + X_{ij}) = \det(A) + \delta_{ij}$, where $\delta_{ij}$ is the $(i, j)$-cofactor of the matrix $A$. Since $A$ has the property $(\mathcal{P})$, it results that $\delta_{ij} = \delta_{ji}$, for any $i, j \in \{1, 2, \dots, n\}$. Hence the adjugate matrix $A^*$ is symmetric. Let us denote $d = \det(A) \neq 0$. We have $AA^* = dI_n = (dI_n)^T = (A^*A)^T = A^T(A^*)^T = A^T A^*$. Since $d \neq 0$ and $d\det(A^*) = \det(AA^*) = \det(dI_n) = d^n$, we deduce that $A^*$ is invertible. Therefore, from the relation $AA^* = A^T A^*$, we obtain $A = A^T$.

b) Consider the matrix $A = (a_{ij})_{1 \le i,j \le n} \in \mathcal{M}_n(\mathbb{C})$ with $a_{11} = a_{12} = 1$ and $a_{ij} = 0$ elsewhere. Since $n \ge 3$, the matrix $A + X_{ij}$ has at least one null row, so $\det(A + X_{ij}) = 0$, for any $i, j \in \{1, 2, \dots, n\}$. Thus, $A$ has the property $(\mathcal{P})$, but $A \neq A^T$.