If you rapidly cool the drink, you're definitely right - energy would flow PCM -> drink. Otherwise, it's semantic if GP is correct. There would be dynamic equilibrium, i.e. flow of thermal energy in both directions, but no true flow from PCM->drink.

The steps would be:

1) Drink > 140, PCM solid

2) energy flows drink -> PCM, drink cools toward 140, PCM liquifies gradually

Then two possibilities:

3a) drink cools to 140-epsilon before PCM liquifies fully

4a) PCM gives energy to drink in dynamic equilibrium, while also losing energy to environment, solidifying

5a) PCM is entirely solid at 140

6a) PCM drops below 140. drink gives energy to PCM and PCM to environment. drink -> PCM thermal conductivity is presumably much higher than PCM -> env, so drink and PCM remain at same temp

OR

3b) PCM liquifies fully before drink hits 140-epsilon

4b) drink and PCM stay at thermal equilibrium (see 6a) while cooling toward 140. energy flows drink -> PCM and PCM -> environment. The former is faster, so the PCM continues liquifying

5b) PCM is entirely liquid at 140. Due to thermal equilibrium, drink is also at 140.

6b) drink stays at 140 while PCM solidifies

7b) see 5a

8b) see 6a

You seem to assume that the PCM encapsulates the drink on all sides. But the cup has a lid, which doesn't have a PCM inside, and which isn't perfectly isolated.

So there will be energy going from the drink to the environment through the lid, which in turn allows energy to flow back from PCM -> drink.