Not sure what to do here. The proof part is very simple you change the respect ${}^{n}C_{r}$ and ${}^{n-1}C_{r-1}$ into fraction form.
I tried in the next part to expand using the $$ P_r $$ equation that they gave and then factoring $x \, (1-x)$ out but I'm not sure how to use the equation that I had proved in answering the question.
The binomial coefficient is defined by $$\binom{n}{k} = \frac{n!}{k! \, (n-k)!}.$$ Now, $$k \, \binom{n}{k} = \frac{n!}{(k-1)! \, (n-k)!} = n \, \frac{(n-1)!}{(k-1)! \, ((n-1) - (k-1))!} = n \, \binom{n-1}{k-1}. $$ The remainder is to follow.
Using the binomial expansion $$(a + b)^{n} = \sum_{k=0}^{n} \binom{n}{k} \, a^{n-k} \, b^{k}$$ then $$ \sum_{r=0}^{n} P_{r} = \sum_{r=0}^{n} \binom{n}{r} \, x^{r} \, (1-x)^{n-r} = (x + 1-x)^{n} = 1^{n} = 1.$$
For the second consider: \begin{align} \sum_{k=1}^{n} k \, P_{k} &= \sum_{k=1}^{n} k \binom{n}{k} \, x^{k} \, (1-x)^{n-k} \\ &= n \, \sum_{k=1}^{n} \binom{n-1}{k-1} \, x^{k} \, (1-x)^{n-k} \\ &= n \, \sum_{k=0}^{n-1} \binom{n-1}{k} \, x^{k+1} \, (1-x)^{n-1-k} \\ &= n \, x \, ((1-x) + x)^{n-1} = n \, x. \end{align}