= Solution
The <holomorphic normal bundle> is the quotient $N_V=T^{1,0}M|_V/T^{1,0}V$, of rank $r$. The tangent sequence is a <short exact sequence> of <holomorphic vector bundles>:
$$
0\longrightarrow T^{1,0}V\longrightarrow T^{1,0}M|_V\longrightarrow N_V\longrightarrow0.
$$
Taking <determinant line bundles> gives $\det(T^{1,0}M)|_V\cong\det(T^{1,0}V)\otimes\det N_V$. Dualizing and rearranging proves <adjunction for a smooth submanifold>:
$$
\boxed{K_V\cong K_M|_V\otimes\det N_V
=K_M|_V\otimes\bigwedge^rN_V.}
$$
The <determinant> identity follows locally by adjoining lifts of a quotient frame to a subbundle frame; changing the lifts adds only off-diagonal blocks, so does not change the <determinant>.
For a smooth divisor, choose reduced local defining functions $f_i$, with $f_i=u_{ij}f_j$ and $u_{ij}$ holomorphic and nowhere zero. The <holomorphic line bundle associated to a divisor> $[V]=\mathcal O_M(V)$ has local frames $e_i=f_i^{-1}$, with $e_j=u_{ij}e_i$. The products $f_ie_i$ define its canonical section vanishing on $V$. Along $V$,
$$
df_i=u_{ij}\,df_j.
$$
Thus $v\mapsto df_i(v)e_i$ patches to a map $T^{1,0}M|_V\to[V]|_V$, whose kernel is $T^{1,0}V$ and which is surjective because each reduced defining function has nonzero differential there. Hence \b[$N_V\cong[V]|_V$]. This is the <normal bundle of a smooth analytic hypersurface>. The divisor construction uses a closed hypersurface: an arbitrary nonclosed embedded hypersurface need not be a divisor. For example, a punctured line in $\mathbb C^2$ cannot be the support of a divisor, since its missing limit point would also lie in every local holomorphic zero set containing that line.
Let $s$ be the section of $E$ defining $V$. The intended hypothesis is that it is a <regular zero locus>: $ds$ has rank $r$ on $V$. In a local trivialization, differentiation of the component functions defines
$$
ds:T^{1,0}M|_V\longrightarrow E|_V.
$$
This is intrinsic: differentiating a change-of-frame matrix introduces terms multiplied by $s$, which vanish on $V$. Its kernel is $T^{1,0}V$, so it induces an isomorphism of <holomorphic normal bundles>. Consequently
$$
\boxed{N_V\cong E|_V,\qquad
K_V\cong\left(K_M\otimes L_1\otimes\cdots\otimes L_r\right)|_V.}
$$
This is the <normal bundle of a regular zero locus>. Here $\det E=L_1\otimes\cdots\otimes L_r$.
Regularity is essential if “vanishing” is read only as equality of underlying sets. On $\mathbb P^2$, the section $Z_0^2$ of $\mathcal O(2)$ vanishes set-theoretically on a smooth line $V$. Its derivative vanishes on $V$, and
$$
N_V\cong\mathcal O_{\mathbb P^1}(1)\not\cong\mathcal O_{\mathbb P^1}(2)=E|_V.
$$
Thus smoothness of the underlying zero set alone does not establish the claimed normal-bundle isomorphism. The preceding proof applies when the equations define the reduced smooth submanifold, equivalently are transverse to zero.
For <Complex projective space>, put $\mathcal O(a)=[aH]$. At a line $\ell\subset\mathbb C^{n+1}$, the tangent space is $\operatorname{Hom}(\ell,\mathbb C^{n+1}/\ell)$. Quotienting $\operatorname{Hom}(\ell,\mathbb C^{n+1})$ by its scalar maps into $\ell$ yields the <Euler sequence>
$$
0\longrightarrow\mathcal O\longrightarrow
\mathcal O(1)^{\oplus(n+1)}\longrightarrow T^{1,0}\mathbb P^n\longrightarrow0.
$$
Its <determinant> gives \b[$K_{\mathbb P^n}\cong\mathcal O(-n-1)=[-(n+1)H]$]. For a smooth complete intersection defined regularly by the homogeneous equations, apply the normal-bundle result to $E=\bigoplus_i\mathcal O(d_i)$:
$$
\boxed{K_V\cong\mathcal O\!\left(\sum_{i=1}^r d_i-n-1\right)|_V
=\left[\left(\sum_i d_i-n-1\right)H\right]|_V.}
$$
This is the <canonical bundle of a regular projective complete intersection>. The same regularity qualification is needed here: the squared-line example would otherwise predict $\mathcal O_{\mathbb P^1}(-1)$ for its <canonical bundle>, whereas the line has $K_V=\mathcal O_{\mathbb P^1}(-2)$.
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