= Solution
The drawn component $L_1$ is an <unknot>. Its exterior is a <solid torus> $V$, in which $L_2$ is a <pattern of a satellite knot>. A <meridian of a solid torus> of $V$ is $\ell_1$, whereas its longitudinal direction is $m_1$. Thus the prescribed gluing sends these directions to $m$ and $\ell$, respectively. This identifies $V$ with a <tubular neighborhood> of $K$ with its zero <Seifert framing>.
Removing the pattern before making this identification gives
$$
Y\cong S^3\setminus\operatorname{int}\nu(P(K)).
$$
Equivalently, meridionally filling the remaining boundary first removes $L_2$ from the construction and leaves $E(K)\cup V=S^3$; its filling core is the required <satellite knot> $C(K)$.
In the original <link diagram>, the two parallel passages through a meridional disk of $V$ run in the same direction; closing them uses a single interchange. The <winding number of a satellite pattern> is therefore two. Untwisting the surrounding disk shows that $P(U)$ is the <unknot> (the pattern is the $(2,1)$ <cable knot>, up to the harmless sign of its single interchange). In particular $\Delta_{P(U)}=1$. The <Satellite formula for the Alexander polynomial> now gives the concise answer
$$
\boxed{\Delta_{C(K)}(t)\doteq\Delta_K(t^2).}
$$
Here $\doteq$ allows multiplication by $\pm t^j$. Reversing the winding orientation replaces $t^2$ by $t^{-2}$, giving the same <Alexander polynomial of a knot> up to such a unit.
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