A state consisting of $n$ identical bosons labeled $1$ , $2$ , ..., $n$ must satisfy
$\begin{align*}P | \psi\left(1,2,...,n \right) \rangle = |\psi\left(1,2,...,n \right) \rangle\end{align*}$
where $P$ is an operator that permutes the $n$ particles. This defines a symmetric state.

A state consisting of $n$ identical fermions labeled $1$ , $2$ , ..., $n$ must satisfy
$\begin{align*}P | \psi\left(1,2,...,n \right) \rangle = \left\{\begin{array}{l}|\psi\left(1,2,...,n \right) \rangle\qquad\te...$
where $P$ is again an operator that permutes the $n$ particles. This defines an antisymmetric state.

It is only possible to antisymmetrize a collection of single-particle states when all of the single-particle states are distinct -- this is the statement of the Pauli exclusion principle. The Pauli exclusion principle applies only to multi-fermion states i.e. it is possible to symmetrize a collection of single-particle states regardless of whether there is redundancy.
Collecting all of this information,
\begin{itemize}
\item Bosons have symmetric wavefunctions.
\item Fermions have antisymmetric wavefunctions.
\item Bosons do not obey the Pauli exclusion principle.
\item Fermions do obey the Pauli exclusion.
\end{itemize}
Therefore, answer (D) is correct.

Solution to 2008 Problem 72