Let $K = P \ \# \ Q = K_1 \ \# \ K_2$ be a knot, where $P$ is a prime knot and "#" denotes connected sum of knots. Let $B$ be a ball in $S^3$ such that $B$ separates $P$ from $Q$, i.e., $B \cap K$ is an arc $\alpha$ and $\partial B \cap K$ consists of two points such that union of $\alpha$ with any arc joining the two points will be the knot $P$. Similarly, let $S$ be a 2-sphere which separates $K_1$ and $K_2$.
Now, suppose $A =S \cap B$ is an annulus such that it forms a boundary for regular neighbourhood of the arc $\alpha$ in $B$. $\partial A$, which is a union of two circles forms another annulus $A^\prime$ on $\partial B$.
How can we say that $A \cup A^\prime$ is a torus? Also, how is it peripheral to $P$ and swallows(contains) $Q$ ? Let $M$ be the area bounded by $A \cup A^ \prime$. Is $S^3 \setminus M$ a solid torus? If not, what is it?
My ideas: I see why $A \cup A^\prime$ is a torus when the arc $\alpha$ is trivial, since it is nothing but boundary of a ball with a cylindrical hole in it. I also think that when $\alpha$ is non-trivial $A \cup A^\prime$ is nothing but the knotted torus.
You can see that $A \cup A'$ is a torus by applying the classification theorem for surfaces: any surface obtained from two annuli by homeomorphically identifying their boundaries is oriented and has Euler characteristic zero, hence is homeomorphic to the torus.
To see that this torus is peripheral to $P$, let $N(P)$ be the regular neighborhood that you originally chose so that $\partial(N(P)) \cap B = A$. In the 3-sphere, you can isotope the annulus $A'$ past infinity to $\partial(N(P))-A$, by an isotopy which is stationary on $A$ and on $P$.
I'm not sure what you mean by saying "Let $M$ be the area bounded by $A \cup A'$." Perhaps what you mean is that $M$ is the bounded component of $\mathbb{R}^3 - (A \cup A')$? If so, then $S^3-M$ goes to $N(P)$ under the isotopy of $S^3$ that I have described, and so yes, $S^3 - M$ is indeed a solid torus.