If $M$ is a semifinite von Neumann algebraactiong on a Hilbert space $H$, a sequence of vectors $\{x_n\}$ is said to converge to $x$ weakly relative to $M$ if $\{x_n\}$ is bounded and $Px_n\rightarrow Px$ for every finite projection $P\in M$.
If $M=B(H)$,the above convergence is the classical weak convergence of vectors.
Let's recall the definition of classical weak convergence of vectors. $\{x_n\}\subset H$ is said to converge weakly to $x$ if $\langle x_n,y\rangle \rightarrow \langle x,y\rangle$ $(n\rightarrow \infty)$ for any $y\in H$.
How to check that the two convergence "coincide" when $M=B(H)$?
The definition, as stated, makes no sense: $Px_n\to Px$ for all projections is trivially equivalent with $x_n\to x$. The way it usually appears in the literature includes the requirement that the projections $P$ are finite in $M$. In the case where $M=B(H)$, this means that we are considering all finite-rank projections.
If $Px_n\to Px$ for all projections $P$, then given $y\in H$ we can choose a rank-one projection $P$ so that $Py=y$. Then $$ \langle (x_n-x),y\rangle=\langle (x_n-x),Py\rangle=\langle P(x_n-x),y\rangle\leq\|y\|\,\|Px_n-Px\|\to0. $$
Conversely, suppose (without loss of generality) that $x_n\to 0$ weakly, that $\|x_n\|\leq c$ for all $n$, and let $P$ be a finite-rank projection. Let $e_1,\ldots,e_m$ be an orthonormal basis of the range of $P$. Then $$ Px_n=\sum_{k=1}^m \alpha_{n,k}e_k, $$ and $|\alpha_{n,k}|\leq c$ for all $n,k$. Then \begin{align} \|Px_n\|^2&=\langle x_n,Px_n\rangle=\sum_{k=1}^m\alpha_{n,k}\,\langle x_n,e_k\rangle\leq \sum_{k=1}^m|\alpha_{n,k}|\,|\langle x_n,e_k\rangle|\\[0.3cm] &\leq c\,\max\{|\langle x_n,e_k\rangle|:\ k=1,\ldots,m\}\to0 \end{align}