Deterministic case: the greedy action $a_g$ (discrete) yield by the greedy policy $\pi_g$ is:
$$\pi_g(s) = a_g = \arg\max_{a\in\mathcal A} q^\pi(s,a)$$
For continuous actions, the $\arg\max_a$ is infeasible to compute. One can thus learn a deterministic policy $\pi_d$, according to the deterministic policy gradient theorem, such that $a_d=\pi_d(s)$ achieves the maximum action-value, i.e. $q^\pi(s,a_d)\ge q^\pi(s,a),\ \forall a\in\mathcal A$.
Now, for both discrete ($a_g$) and continuous ($a_d$) actions these are greedy, in the sense that their respective policies should assign a probability of one to those (if viewed as if being stochastic), and zero to all sub-optimal actions:
$$
\pi_g(a\mid s) = \begin{cases}1 & \text{if $a=a_g$} \\ 0 & \text{otherwise} \end{cases},
$$
similarly for $\pi_d$ and $a_d$. This means that $\pi_g$ (viewed as stochastic) samples the greedy action all the times, so it always exploits.
Stochastic case: to simplify let's assume the policy is Gaussian, i.e. $\pi(a\mid s) = \mathcal{N}(\mu,\sigma^2\mid s)$. I assume $\mu$ and $\sigma$ to be meaningful, like being learned to maximize the action-value function, $q^\pi$. If so, in my opinion, the most likely action (i.e. the distribution's mode) should have the largest action-value. In the case of Gaussian distributions, the mean $\mu$ is the most likely action (being also its mode) and so it should be the greedy action, i.e. $a_g=\mu$ such that $q^\pi(s,\mu)\ge q^\pi(s,a),\forall a \in\mathcal A$.
Therefore $\pi$ assigns the following probabilities:
$$
\pi(a\mid s) = \begin{cases}\alpha e^{-\frac12} & \text{if $a=\mu$} \\
\alpha e^{-\frac12\big(\frac{a-\mu}{\sigma}\big)^2} & \text{otherwise} \end{cases},
$$
where $\alpha = (\sigma\sqrt{2\pi})^{-1}$. Now, from this we can see that the considered stochastic policy is greedy $\alpha e^{-\frac12}$ of the times (since is the probability of sampling the mean $\mu$ as action.) Since there always exist a deterministic policy (see slide 41) we can, in principle, turn the stochastic $\pi$ into a deterministic policy $\pi_d$ such that:
$$
\pi_d(a\mid s) = \begin{cases}1 & \text{if $a=\mu$} \\ 0 & \text{otherwise} \end{cases},
$$
in order to be greedy all the times: with probability one, $\mu=\pi_d(s)$ always outputs the mean $\mu$ (action that maximizes $q^\pi$) in state $s$ - indeed, the value of $\mu$ is state-dependent, meaning that a different state could have a different mean.
(Disclaimer: I don't know if the above is mathematically sound and correct but that's my intuition. So, please correct me eventually.)