Put $D=\sum_x d(x)$. Then the probability measure $\mu(x)\colon=d(x)/D$ is reversible for the Markov chain in exam. In particular the normalized counting measure is a reversible probability measure for the random walk on a regular undirected graph.
Since $x\sim y$ iff $y\sim x$, we conclude that $\{(x,y):x\in S,y\sim x\}=\{(x,y):y\in S,x\sim y\}$ and thus for all $f,g\in B(S)$: $$ \sum_x Pf(x)g(x)\mu(x) =\sum_x\Big(\frac1{d(x)}\sum_{y\sim x}f(y)\Big)g(x)\frac{d(x)}D =\frac1D\sum_x\sum_{y\sim x}g(x)f(y) =\frac1D\sum_y\sum_{x\sim y}f(y)g(x) =\sum_y f(y)Pg(y)\mu(y)~. $$