Calculating Vmax (or Kcat) given specific activity of an enzyme over length of time incubated with an activator
The enzyme catalysis reaction following MM kinetics can be depicted as:

$$\Large\ce{E + S&lt;=&gt;[k_f][k_r]ES-&gt;[k_{cat}]E + P }$$

Note that $V_{max} = k_{cat}.[E_0]$ i.e. conversion rate constant (turnover number) times the total amount of enzyme.

You should note that the enzyme here is p38 and substrate is MSK1 (which in turn is an enzyme). When MSK1 gets phosphorylated, it assumes an active form. MSK1 activity (and specific activity) would reflect the concentration of phosphorylated MSK1 (MSK1-P) i.e. the product. Note that the $k^{^{MSK1-P}}_{cat}$ is its intrinsic property and it does not change; only the $[E^{^{MSK1-P}}_0]$ changes. 

I hope your doubt is clear.