= Solution
\b[The divisor determines the bundle through its local equations.] Choose an <open cover> on which $V$ has a reduced <holomorphic> defining equation $f_i$, using $f_i=1$ on sets disjoint from $V$. Because $V$ is a <complex submanifold> of <codimension of a submanifold> one, its ideal is locally generated by a coordinate; consequently $f_i/f_j$ is a nowhere-zero <holomorphic function> on each overlap. Inside the <sheaf of meromorphic functions on a complex manifold>, consider the <locally free sheaf> of rank one
$$
\mathcal O_M(V)|_{U_i}=f_i^{-1}\mathcal O_{U_i}.
$$
These local descriptions agree on overlaps. With $e_i=1/f_i$, its <transition functions of a vector bundle> are specified by
$$
e_i=\frac{f_j}{f_i}e_j.
$$
The ratios satisfy the <Čech cocycle condition>, and the associated <holomorphic line bundle associated to a divisor> is $[V]$. The section $1=f_i e_i$ is <holomorphic> and vanishes to order one along $V$. Replacing $f_i$ by $u_i f_i$, with $u_i$ a <holomorphic> unit, replaces the frame by $u_i^{-1}e_i$ without changing the subsheaf of meromorphic functions. Refining the cover changes no local sections either. This proves \b[choice independence up to isomorphism] rather than merely producing a bundle for one cover.
For a <complex manifold> of dimension $n$, the <canonical bundle> is
$$
\boxed{K_M=\bigwedge\nolimits^n(T^{1,0}M)^*.}
$$
It is the <holomorphic line bundle> of holomorphic top-degree <differential forms>. To compute it on <Complex projective space>, let $S=\mathcal O(-1)$ be the <complex tautological line bundle>, so that $S^*=[H]=\mathcal O(1)$ is the <hyperplane line bundle>. At a line $\ell\subset\mathbb C^{n+1}$, the <tangent space> is $\operatorname{Hom}(\ell,\mathbb C^{n+1}/\ell)$. Tensoring the tautological quotient sequence by $S^*$ gives the <Euler sequence on complex projective space>
$$
0\longrightarrow\mathcal O\longrightarrow
\mathcal O(1)^{\oplus(n+1)}
\longrightarrow T^{1,0}\mathbb P^n
\longrightarrow0.
$$
Taking the top <exterior power> gives $\det T^{1,0}\mathbb P^n=\mathcal O(n+1)$; taking its <dual bundle> yields
$$
\boxed{K_{\mathbb P^n}\cong[H]^{\otimes(-n-1)}.}
$$
Here a negative <tensor power> means the corresponding positive <tensor power> of the <dual bundle>.
The proposed sections over $U$ form a <sheaf>: compatible maps into $L$ glue uniquely, and their projections remain $\phi$. Addition and multiplication by <holomorphic functions> are defined in the fibres of $L$, making this a sheaf of $\mathcal O_X$-modules. If $e_i$ is a <holomorphic local frame> of $L$ on $W_i\subset Y$, every section over $U\subset\phi^{-1}(W_i)$ has the unique expression
$$
\theta(x)=a(x)e_i(\phi(x)),\qquad a\in\mathcal O_X(U).
$$
Thus the sheaf is a <locally free sheaf> of rank one. Its <transition functions of a vector bundle> are those of $L$ composed with $\phi$, so the corresponding <pullback vector bundle> is precisely $\phi^*L$. This proves \b[invertibility locally and compatibility globally].
For the local <blowup of a complex manifold at a point>, work on the projective chart $Y_i\ne0$ and write
$$
u=x_i,\qquad t_j=Y_j/Y_i\quad(j\ne i).
$$
All the incidence equations reduce to $x_j=ut_j$. Therefore $(u,(t_j)_{j\ne i})$ gives a <holomorphic coordinate> chart isomorphic to $\mathbb C^n$. On an overlap with $Y_\ell\ne0$,
$$
u'=ut_\ell,\qquad t'_i=1/t_\ell,\qquad
t'_j=t_j/t_\ell\quad(j\ne i,\ell),
$$
which is a <biholomorphism> where $t_\ell\ne0$. The incidence subset is closed in the product of two Hausdorff spaces and is covered by these charts; hence it is a <complex manifold> of dimension $n$. In each chart the exceptional fibre is $u=0$, so it is a smooth <complex submanifold>, with its induced projective coordinates giving
$$
\boxed{E\cong\mathbb P^{n-1}.}
$$
Away from the origin the inverse of the projection is $x\mapsto(x,[x])$, which is <holomorphic>; the projection is therefore a <biholomorphism> outside $E$.
To find the <canonical bundle>, pull back the nowhere-zero top form on $\mathbb C^n$. In the chart above, its <Jacobian determinant> gives
$$
\pi_1^*(dx_1\wedge\cdots\wedge dx_n)
=\pm u^{n-1}\,du\wedge\bigwedge_{j\ne i}dt_j.
$$
The sign depends only on the ordering of the coordinates. This is a section of $K_X$ with <divisor on a complex manifold> exactly $(n-1)E$: it has that vanishing order in every exceptional chart and no zeros elsewhere. A nonzero <meromorphic section of a holomorphic line bundle> with divisor $D$ identifies its <holomorphic line bundle> with $[D]$, by sending the local generator $1/f_i$ of $\mathcal O(D)$ to the nowhere-zero frame obtained by dividing the section by $f_i$. Consequently
$$
\boxed{K_X\cong[E]^{\otimes(n-1)}.}
$$
This also includes $n=1$, when the local map is an isomorphism and the exponent is zero.
For a general <complex manifold>, choose a coordinate neighbourhood $U$ of $P$, with $P$ sent to $0$, and replace $U$ by the inverse image of this coordinate neighbourhood in the local <blowup of a complex manifold at a point>. Glue this space to $M\setminus\{P\}$ along $U\setminus\{P\}$ using the preceding <biholomorphism>. The resulting charts give a <complex manifold> $\widetilde M$ and a <holomorphic map> $\phi$; the exceptional fibre is $\mathbb P^{n-1}$ and the map is a <biholomorphism> off that fibre. For completeness, the local projection is proper because its inverse image over a compact subset is closed in that subset times compact <Complex projective space>. This ensures the gluing is Hausdorff: separate points with distinct images downstairs, and points over $P$ inside the local blowup. The atlas is second countable as well.
The top <exterior power> of the differential defines a global <holomorphic bundle map> $\phi^*K_M\to K_{\widetilde M}$. Equivalently, it is a section of $K_{\widetilde M}\otimes(\phi^*K_M)^{-1}$. The same local <Jacobian determinant> has vanishing order $n-1$ on $E$, and the differential is invertible elsewhere. Applying the <holomorphic line bundle associated to a divisor> construction gives the <canonical bundle formula for a point blowup>
$$
\boxed{K_{\widetilde M}\cong\phi^*K_M\otimes[E]^{\otimes(n-1)}.}
$$
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