Why aren't these two solutions equivalent? Combinatorics problem

Your second solution is the correct one.

Your first solution is incorrect because you overcount the scenarios where a professor is picked in the second step.

The outcome where you pick the first three professors in the first step followed by the fourth professor in the second step: $\{p_1,p_2,p_3\},\{p_4,s_1\}$ is also counted where you picked the last three professors in the first step and the first professor in the second step: $\{p_2,p_3,p_4\},\{p_1,s_1\}$. These outcomes should be considered the same however since in both scenarios you have the same five people selected.

Be careful not to overcount things with multiplication principle. Objects selected in one step are treated differently than objects selected in a later step.

The first answer is wrong. It overestimates the count by double-counting the four-professors solutions. This is because each can begin with three of the four in four different ways. Note that $$\binom{4}{3}\binom{7}{2}+4\binom{4}{4}\binom{7}{1}=112.$$Although "double counting" referred above to a fallacy, it's also the name of a valid, useful technique one should be happy to use.