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Here is a proof that all such embeddings are knotless. Consider the four half-planes bounded by $A$ which each contain one of the points $b_i$. Then these planes give an open-book decomposition which contains $K_{4,4}$. Each plane contains four edges, but if $L$ is a cycle of $K_{4,4}$, then it can only contain at most two edges from each plane, since it must pass through $b_i$. This gives an arc presentation of $L$ which has arc-index at most $4$. But it is known that every non-trivial knot has arc index at least $5$. Therefore the embedding of $K_{4,4}$ is knotless.

Below is my original answer, giving one explicit knotless embedding.

Let the points on $A$ be $\{(-4,0,0), (-1,0,0), (1,0,0), (4,0,0)\}$. Let the points on $B$ be $\{(0,-3,1), (0,-2,1), (0,2,1), (0,3,1)\}$. Look at the projection $\pi$ of $K_{4,4}$ onto the $xy$ plane, which has exactly four crossings. A non-trivial knot has at least three crossings in any diagram.

Suppose $L$ is a non-trivial knot or link embedded in this $K_{4,4}$. Then $L$ has at most 8 edges. If $\pi(L)$ contains all four crossings, then $L$ is a pair of unlinked circles. If $\pi(L)$ contains three crossings, then by symmetry, it doesn't matter which three are chosen, and $L$ is an unknot. Therefore this embedding is knotless. There is however a Hopf link in this $K_{4,4}$, which uses two opposite crossings, so this embedding is not linkless.

Different choices of points could give many more crossings in the projection, and I wouldn't like to try to extend this argument to anything much more complicated.

Here is a proof that all such embeddings are knotless. Consider the four half-planes bounded by $A$ which each contain one of the points $b_i$. Then these planes give an open-book decomposition which contains $K_{4,4}$. Each plane contains four edges, but if $L$ is a cycle of $K_{4,4}$, then it can only contain at most two edges from each plane, since it can contain at most two of the edges which meet each $b_i$. This gives an arc presentation of $L$ which has arc-index at most $4$. But it is known that every non-trivial knot has arc index at least $5$. Therefore the embedding of $K_{4,4}$ is knotless.

Below is my original answer, giving one explicit knotless embedding.

Let the points on $A$ be $\{(-4,0,0), (-1,0,0), (1,0,0), (4,0,0)\}$. Let the points on $B$ be $\{(0,-3,1), (0,-2,1), (0,2,1), (0,3,1)\}$. Look at the projection $\pi$ of $K_{4,4}$ onto the $xy$ plane, which has exactly four crossings. A non-trivial knot has at least three crossings in any diagram.

Suppose $L$ is a non-trivial knot or link embedded in this $K_{4,4}$. Then $L$ has at most 8 edges. If $\pi(L)$ contains all four crossings, then $L$ is a pair of unlinked circles. If $\pi(L)$ contains three crossings, then by symmetry, it doesn't matter which three are chosen, and $L$ is an unknot. Therefore this embedding is knotless. There is however a Hopf link in this $K_{4,4}$, which uses two opposite crossings, so this embedding is not linkless.

Different choices of points could give many more crossings in the projection, and I wouldn't like to try to extend this argument to anything much more complicated.

I think it might depend on the choice of points. Here is one example of a knotless embedding. Let the points on $A$ be $\{(-4,0,0), (-1,0,0), (1,0,0), (4,0,0)\}$. Let the points on $B$ be $\{(0,-3,1), (0,-2,1), (0,2,1), (0,3,1)\}$. Look at the projection $\pi$ of $K_{4,4}$ onto the $xy$ plane, which has exactly four crossings. A non-trivial knot has at least three crossings in any diagram.

Suppose $L$ is a non-trivial knot or link embedded in this $K_{4,4}$. Then $L$ has at most 8 edges. If $\pi(L)$ contains all four crossings, then $L$ is a pair of unlinked circles. If $\pi(L)$ contains three crossings, then by symmetry, it doesn't matter which three are chosen, and $L$ is an unknot. Therefore this embedding is knotless. There is however a Hopf link in this $K_{4,4}$, which uses two opposite crossings, so this embedding is not linkless.

Different choices of points could give many more crossings in the projection, and I wouldn't like to try to extend this argument to anything much more complicated.

Here is a proof that all such embeddings are knotless. Consider the four half-planes bounded by $A$ which each contain one of the points $b_i$. Then these planes give an open-book decomposition which contains $K_{4,4}$. Each plane contains four edges, but if $L$ is a cycle of $K_{4,4}$, then it can only contain at most two edges from each plane, since it must pass through $b_i$. This gives an arc presentation of $L$ which has arc-index at most $4$. But it is known that every non-trivial knot has arc index at least $5$. Therefore the embedding of $K_{4,4}$ is knotless.

Below is my original answer, giving one explicit knotless embedding. 

Let the points on $A$ be $\{(-4,0,0), (-1,0,0), (1,0,0), (4,0,0)\}$. Let the points on $B$ be $\{(0,-3,1), (0,-2,1), (0,2,1), (0,3,1)\}$. Look at the projection $\pi$ of $K_{4,4}$ onto the $xy$ plane, which has exactly four crossings. A non-trivial knot has at least three crossings in any diagram.

Suppose $L$ is a non-trivial knot or link embedded in this $K_{4,4}$. Then $L$ has at most 8 edges. If $\pi(L)$ contains all four crossings, then $L$ is a pair of unlinked circles. If $\pi(L)$ contains three crossings, then by symmetry, it doesn't matter which three are chosen, and $L$ is an unknot. Therefore this embedding is knotless. There is however a Hopf link in this $K_{4,4}$, which uses two opposite crossings, so this embedding is not linkless.

Different choices of points could give many more crossings in the projection, and I wouldn't like to try to extend this argument to anything much more complicated.

I think it might depend on the choice of points. Here is one example of a knotless embedding. Let the points of $A$ be $\{(-4,0,0), (-1,0,0), (1,0,0), (4,0,0)\}$. Let the points of $B$ be $\{(0,-3,1), (0,-2,1), (0,2,1), (0,3,1)\}$. Look at the projection $\pi$ of $K_{4,4}$ onto the $x$-$y$ plane, which has exactly four crossings. A non-trivial knot has at least three crossings in any diagram.

Suppose $L$ is a non-trivial knot or link embedded in $K_{4,4}$. If $\pi(L)$ contains all four crossings, then $L$ is a pair of unlinked circles. If $\pi(L)$ contains three crossings, then by symmetry, it doesn't matter which three are chosen, and $L$ is an unknot. Therefore this embedding is knotless. There is however a Hopf link in this $K_{4,4}$, which uses two opposite crossings, so this embedding is not linkless.

Different choices of points could give many more crossings in the projection, and I wouldn't like to try to extend this argument to anything much more complicated. Note, there are also choices of points where the $K_{4,4}$ is not embedded.

I think it might depend on the choice of points. Here is one example of a knotless embedding. Let the points on $A$ be $\{(-4,0,0), (-1,0,0), (1,0,0), (4,0,0)\}$. Let the points on $B$ be $\{(0,-3,1), (0,-2,1), (0,2,1), (0,3,1)\}$. Look at the projection $\pi$ of $K_{4,4}$ onto the $xy$ plane, which has exactly four crossings. A non-trivial knot has at least three crossings in any diagram.

Suppose $L$ is a non-trivial knot or link embedded in this $K_{4,4}$. Then $L$ has at most 8 edges. If $\pi(L)$ contains all four crossings, then $L$ is a pair of unlinked circles. If $\pi(L)$ contains three crossings, then by symmetry, it doesn't matter which three are chosen, and $L$ is an unknot. Therefore this embedding is knotless. There is however a Hopf link in this $K_{4,4}$, which uses two opposite crossings, so this embedding is not linkless.

Different choices of points could give many more crossings in the projection, and I wouldn't like to try to extend this argument to anything much more complicated.