For hermitian $B$, the inequality is very easy to prove: use that $\|B\| - B$ is positive semidefinite together with the fact that the product of two positive semidefinite matrices has nonnegative trace. This gives $\mathrm{tr}(A(\|B\| - B)) \geq 0$, from which the inequality follows immediately.

For generic non-hermitian $B$, the left-hand side of your inequality is not real. So you need to specify more precisely what you mean for general $B$.

For hermitian $B$, the inequality is very easy to prove: use that $\|B\| - B$ is positive semidefinite together with the fact that the product of two positive semidefinite matrices has nonnegative trace. This gives $\mathrm{tr}(A(\|B\| - B)) \geq 0$, from which the inequality follows immediately.

For general $B$, we can therefore argue that

$$\mathrm{Re}(\mathrm{tr}(AB)) = \tfrac{1}{2}\mathrm{tr}(A(B + B^*))\leq \mathrm{tr}(A)\cdot \tfrac{1}{2}\|B + B^*\| \leq \mathrm{tr}(A) \|B\|.$$

Upon multiplying $B$ by a suitable complex scalar, this implies that

$$|\mathrm{tr}(AB)| \leq \mathrm{tr}(A) \|B\|$$

is true for all positive semidefinite $A$ and any $B$.

I don't have the book at hand right now, but I imagine that this is somewhere to be found in the early chapters of Bhatia's Matrix Analysis.