Details of the method mentioned in @Batman's comment:
We can view each square on the chessboard as a vertex on a graph consisting of $64$ vertices, and two vertices are connected by an edge if and only if a knight can move from one square to another by a single legal move.
Since knight can move to any other squares starting from a random square, then the graph is connected (i.e. every pair of vertices is connected by a path).
Now given a vertex $i$ of the graph, let $d_i$ denote the degree of the vertex, which is number of edges connected to the vertex. This is equivalent to number of possible moves that a knight can make at that vertex (square on chessboard). Since the knight moves randomly, transition probabilities from $i$ to its neighbors is $1/d_i$.
Now since the chain is irreducible (since the graph is connected) the stationary distribution of the chain is unique. Let's call this distribution $\pi$. Now we claim the following:
Claim Let $\pi_j$ denote $j^\text{th}$ component of $\pi$. Then $\pi_j$ is proportional to $d_j$.
Proof Let $I$ be the fuction on vertices of the graph such that $I(i)=1$ if $i$ is a neighbor of $j$, and $I(i)=0$ otherwise. Then
$$
d_j=\sum_i I(i)=\sum_i d_i \cdot \frac{I(i)}{d_i} = \sum_i d_i p_{ij}
$$
where $p_{ij}$ is the transition probability from $i$ to $j$. Hence we have $dP=d$ where $P$ is the transition matrix of the chain, and $d=(d_1,\cdots,d_j,\cdots,d_{64})$. Thus $\pi P=\pi \implies$ Claim
Therefore, it follows that after normalising we have
$$
\pi_j=d_j/\sum_i d_i
$$
Finally we recall the following theorem
Theorem If the chain is irreducible and positive recurrent, then
$$
m_i=1/\pi_i
$$
Where $m_i$ is the mean return time of state $i$, and $\pi$ is the unique stationary distribution.
Thus if we call the corner vertex $1$, we have
$$
m_1=1/\pi_1
$$
You can check that $\sum_i d_i = 336$, and we have $d_1=2$ (at corner knight can make at most $2$ legal moves. Therefore $\pi_1=1/168$ and
$$
m_1=168
$$
The stationary distribution is defined as the normalized number of moves from a given position. In symbols, for a given position $x $ it is $\frac {deg (x)}{\sum_{x∈S} deg(x)}$, where $deg $ indicates the number of possible moves from $x $.
For a queen on a chessboard, if it is on any of the $28$ squares adjacent to the outer edge (including corners), there are $21$ possible moves ($7$ ranks, $7$ files and $7$ diagonals). If it is on any of the $20$ squares that are in the second concentric frame (i.e., all squares separated from the outer edge by one square), there are $23$ possible moves (because there are two additional diagonal moves).
If it is on any of the $12$ squares that are in the third concentric frame (i.e., all squares separated from the outer edge by two squares), there are $25$ possible moves (because there are two further additional diagonal moves). Lastly, if the queen is on one of the 4 central squares, there are $27$ possible moves (again two further additional diagonal moves).
So we have for the corner $deg (x)=21$ and for the total chessboard $$\sum_{x∈S} deg(x)= 28 \cdot 21 + 20 \cdot 23 + 12 \cdot 25 + 4 \cdot 27 = 1456$$
which leads to an expected number of moves of $1456 /21\approx 69.3$ to return to the corner.
Note that, in my opinion, the values of $1452$ (instead of $1456$) and the resulting $69.14$ (instead of $69.3$), both provided in the solutions that you cite, might be the result of a typo (the value of $1456$ is well established for problems on Queen random walks).
Best Answer
Consider a graph whose nodes are the black squares of the chessboard. At each black square, find the probability with which a consecutive pair of knight moves would land on each other black square. This gives a transition matrix for a Markov process that consists of pairs of knight moves.
In fact this new Markov process is just a way of describing all the even-numbered states of the original process (if you consider the starting state to be state number zero). If your mean return time to the bottom left corner is $168$ in the original process, it is $84$ in the new process.