In this answer I assume given scores are pairwise didtinct. Note that the probability of two scores being equal is 0 since we have continuous probability. I think the same proof can be tweaked to span the case where two probabilities are equal but it will make it more complicated.
Let $p_1, \dots p_n$ be the set of employees sorted in descending order according to the outcome of the first task. Let $X_i$ be an indicator random variable, that the employee $p_i$ wins a prize. This means
$$X_i =
\begin{cases}
1&;\text{$p_i$ wins a prize,}\\
0&;\text{Otherwise.}
\end{cases}$$
Let $C$ be a random variable equals to the number of employees who win a prize. Note that $C = \sum\limits_{i=1}^{n} X_i$ and by linearity of expectation we get $E[C] = \sum\limits_{i=1}^{n}E[X_i]$. Moreover, note that the variables $X_i$ are mutually independent, since an employee $p_i$ wins a prize if and only if the score of $p_i$ in the second task is greater than the score of $p_j$ in the same task for all $j < i$.
It is easy to see that $$E[X_i] = Pr[X_i] = \frac{1}{i}$$ (think about the values of the scores the first $i$ employees get and the probability that the $i$th gets the highest of them). Now we can compute
$$E[C] = \sum\limits_{i=1}^{n}E[X_i]= \sum\limits_{i=1}^n\frac{1}{i} = H_n \leq \ln n,$$
where $H_n$is the $n$th term of the harmonic series.
Now since we already discussed that the variables are independent, we can apply Chernoff bounds to prove that the probability, that the expected value is higher than a constant factor of $\ln n$ is very small and hence, with high probability the expected value is not greater than a constant factor of $\ln n$.
Here is the extension about Chernoff bounds.
Chernoff bound: Let $\mu := E[C]$ and let $\delta < 2e - 1$ be a positive constant, the bound can be written as follows (see link):
$$Pr[C > (1+\delta)\mu] <
e^{-\mu\delta^2/4}.$$
Now set $\delta = 4$. We get
$$Pr[C > 5\lg n] < e^{-16/4\ln n} = \frac{1}{n^{4}}$$
