If a symmetric PSD matrix has a zero entry on the main diagonal, its determinant is zero

Let matrix $A$ be symmetric and positive semidefinite. If there exists $a_{ii} = 0$, then $$\det (A) = 0$$ Here, $a_{ii}$ denotes any diagonal element of $A$.

I am really struggling to get an idea how to tackle this question. Can anyone help?

This is a basic consequence of Hadamard's inequality, which says that for an $n \times n$ positive semidefinite matrix $P$ with entries $(p_{i,j})$ that $$ \det(P) \leq \prod_{j = 1}^n p_{j,j}. $$ In your case we know that the RHS is zero, so the LHS must be as well (positive semidefinite matrices have nonnegative determinant).

If the "Hadamard's inequality" that you know is the version for arbitrary matrices, then we can immediately get this version using the so-called Cholesky decomposition of $\mathbf{A}$.

Here's another way to view it. We have for positive semidefinite $A$ that $\mathrm{det}(A) \neq 0$ iff $A >0$, i.e., it is positive definite with all of its eigenvalues strictly positive. This can be seen via the spectral decomposition, noting that the determinant of $A$ is the product of its eigenvalues.

Now if the $i$-th diagonal element of $A$ is $0$. Then let $e_i$ be the vector with a $1$ in the $i$-th component and zeros elsewhere. Then $e_i^T A e_i = 0$ which by definition implies that $A$ is not positive definite and hence $\mathrm{det}(A) = 0$.