$$\sum_{k=1}^{n}kp_n(k)$$$$=\sum_{k=1}^{n}k \binom{n}{k} D_{n-k}$$$$= \sum_{k=1}^{n} \left (k \frac{n!}{k!} (n-k)! \sum_{i=1}^{n-k} \frac{(-1)^{i+1}}{i!} \right)$$$$=\sum_{k=1}^{n} \frac{n! \sum_{i=1}^{n-k} \frac{(-1)^{i+1}}{i!}}{(k-1)!}$$$$=n! \sum_{k=1}^{n} \frac{1}{e(k-1)!}=\frac{n!}{e} \sum_{k=1}^{n} \frac{1}{(k-1)!}$$$$=n!$$

$$\sum_{k=1}^{n}kp_n(k)$$$$= \sum_{k=1}^{n}k \binom{n}{k} D_{n-k}$$$$= \sum_{k=1}^{n} \left(k \frac{n!}{k!} (n-k)! \sum_{i=1}^{n-k} \frac{(-1)^{i+1}}{i!} \right)$$$$ = \sum_{k=1}^{n} \frac{n! \sum_{i=1}^{n-k} \frac{(-1)^{i+1}}{i!}}{(k-1)!}$$$$=n! \sum_{k=1}^{n} \frac{1}{e(k-1)!}=\frac{n!}{e} \sum_{k=1}^{n} \frac{1}{(k-1)!}$$$$=n!$$

Here is a combinatorial solution. $\sum_{k=1}^nkp_k(n)$ counts the number of ordered pairs $(\pi,j)$, where $\pi$ is a permutation and $j$ is a fixed point of $\pi$. This is because there are $p_k(n)$ permutations with $k$ fixed points, and for each of those, there are $k$ choices for $j$. To count the number of ordered pairs $(\pi,j)$, let us ask how many choices of $\pi$ there are for a given $j$. The number of permutations which have a given number $j$ as a fixed point is $(n-1)!$. Therefore, $$\sum_{k=1}^nkp_k(n)$$=# of $(\pi,j)$ such that $j$ is fixed point of $\pi$ $=\sum_{j=1}^n$ # of $\pi$ such that $j$ is fixed point of $\pi$ $$=\sum_{j=1}^n (n-1)!$$$$= n\cdot (n-1)!=n!.$$ For $(b)$, you are instead counting ordered sequences $(\pi,j_1,j_2,\dots,j_s)$ so that $j_1,\dots,j_s$ are all fixed points of $\pi$, as there are $k^s$ ways to choose the list of $j_1,\dots,j_s$ when $\pi$ has $k$ fixed points. Given a list which contains $i$ distinct values, there are $(n-i)!$ ways to choose a permutation $\pi$ which has all those values as fixed points. The number of lists $j_1,\dots,j_s$ which have exactly $i$ distinct values is $R(s,i)\cdot \frac{n!}{(n-i)!}.$ This is because such a list is formed by partitioning $\{1,2,\dots,s\}$ into $n$ parts, ordering the parts by smallest element, then assigning the first part one of $n$ numbers, then assigning the second one one of $(n-1)$ numbers different than the first, and so on. Therefore, $$ \sum_{k=1}^nk^sp_k(n)$$= # of $(\pi,j_1,\dots,j_s)$ such that each $j_i$ is fixed point of $\pi$ $= \sum_{i=1}^s$ (# $(j_1,\dots,j_s)$ $i$ distinct values) $\cdot$ (# of $\pi$ where $i$ particular points are fixed) $$= \sum_{i=1}^s R(s,i)\cdot \frac{n!}{(n-i)!}\cdot (n-i)!$$$$= n! \sum_{i=1}^s R(s,i).$$