I'm stuck to solve the problems below:
(a) Let $p(n, k)$ be the number of partitions of $n$ into exactly $k$ parts. Show that
$$ \sum_{n, k \geq 0} p(n, k) x^{n} t^{k}=\prod_{i \geq 1} \frac{1}{1-x^{i} t}=\sum_{k \geq 0} \frac{x^{k} t^{k}}{(1-x)\left(1-x^{2}\right) \cdots\left(1-x^{k}\right)} . $$
(b) Find a similar generating function for the number $p_{d}(n, k)$ of partitions of $n$ into exactly $k$ distinct parts
I know the generating function of the partition number: $$ \sum_np(n)x^n=\prod_{i \geq 1}\frac{1}{1-x^i} $$ and how to prove it, but I've never heard about a two-variable generating function.
I did a little research about it, and I got some information about it:
Another way to get a generating function for $p(n, k)$ is to use a two-variable generating function for all partitions, in which we count each partition $\lambda=\left(\lambda_1, \lambda_2, \ldots, \lambda_k\right) \vdash n$ with weight $y^k x^n$, where $n$ is the size and $k$ is the number of parts. The monomial giving the weight contribution for a single part of $i$ now becomes $y x^i$ instead of just $x^i$ and accordingly, we get the generating function $$ P(x, y)=\sum_{n, k} p(n, k) y^k x^n=\prod_{i=1}^{\infty} \frac{1}{1-y x^i}=\frac{1}{(1-y x)\left(1-y x^2\right)\left(1-y x^3\right) \cdots} . $$
(I got it from : https://math.berkeley.edu/~mhaiman/math172-spring10/partitions.pdf)
From the information I got, I cannot understand how I can just change $x^i$ into $yx^i$, and also what the 'weight $yx^i$' means. Also, I have no idea how to start on problem (b).
Please help me.
UPDATE: I thought a little bit more, and I think this is the required answer from (b): $$ \prod_{i \geq 1}(1+yx^i) $$ by the same fashion in the blockquote(if I assume that I can understand that fully). But I'm stuck again to transform this formula into another form like part (a).
This is an elaboration on Haiman's notes that you found.
By "weight," he means the monomial contribution of a partition in the expanded expression for the generating function. For example, just using $x$ terms, the partitions of 4 in the expansion correspond to $$ x^4 + x^3x^1 + x^2x^2 + x^2x^1x^1 + x^1x^1x^1x^1 = 5x^4.$$
As a colleague likes to say in his classes, "you can't stop me from" replacing the $(1-x^i)$ terms with $(1-yx^i)$. What does this $y$ do to the partitions of 4? \begin{gather} yx^4 + (yx^3)(yx^1) + (yx^2)(yx^2) + (yx^2)(yx^1)(yx^1) + (yx^1)(yx^1)(yx^1)(yx^1) \\ = y^1x^4 + 2y^2x^4 + y^3x^4 + y^4x^4. \end{gather} The $y$ terms record the number of parts. We see algebraically that there is 1 partition of 4 with 1 part, 2 partitions of 4 with 2 parts, 1 partition of 4 with 3 parts, and 1 partition of 4 with 4 parts. Note that setting $y=1$ reduces this expression to $5x^4$. (Euler came up with this approach.)
For the transformation from the product to the sum, think about the operation of partition conjugation. If you draw the Ferrers diagram of a partition, where each part corresponds the the number of dots in a row, conjugation reflects along the diagonal and has the effect of swapping rows and columns. Thus $\lambda \in P(n,k)$ with $k$ parts has first column of height $k$, and its conjugate has first row / part $k$. Therefore, $p(n,k)$ counts both partitions with $k$ parts and partitions with largest part $k$.
The infinite sum is over these conjugate partitions. They are grouped by largest part $k$: the numerator insures that $k$ is a part and the $t^k$ indicates that the partition has $k$ columns, while the denominator allows for any number of parts 1 through $k$.
[Graphics idea beyond my MSE skills: In a Ferrers diagram of, say, $(4,3,2)$, put a $yx$ in the 3 boxes of the first column and an $x$ in the other 6 boxes. Conjugating gives $(3,3,2,1)$ where the first row has $yx$ in each of its 3 boxes while the subsequent rows consist of boxes with $x$. The products of the box content for both is $y^3x^9$. In the first partition, that corresponds to a 3-part partition of 9. In the second, it's a partition with largest part 3, the first 3 corresponding to $(yx)^3$.]
You're on the right track for (b). The equivalent infinite sum expression is derived on pp4-5 of Haiman's notes.