Let $f:M\to Gr(N,k)$ be a holomorphic map where $M$ is an algebraic curve(i.e. closed compact Riemann surface) and $Gr(N,k)$ is gramssmanian of $k+1$ plane space in $C^{N+1}$ with $C$ being complex number.
Let $B$ be an $k+1$ plane of $C^{N+1}$ represented by an exterior algebra element. Then from inner product $(-,-)$ on exterior algebra induced from $C^{N+1}$ standard inner product, one can define $\Sigma_B=\{\Lambda\in Gr(N,k)\vert (B,\Lambda)=0\}$ which clearly defines codimension 1 subvariety of $Gr(N,k)$.
"Let $f:M\to Gr(N,k)$ be a holomoprhic curve s.t. the image $f(M)$ does not belong to $\Sigma_B$."
$\textbf{Q:}$ Is it possible always to choose a $B$ s.t. $f(M)\cap\Sigma_B=\emptyset$? $f(M)$ should be at most dimension 1. However $\Sigma_B$ is codimension 1. The intersection is possible to be non-empty from dimension counting of component of intersection dimension$\geq 1+(n-1)-n=0$ where I have assumed $f(M)$ smooth image. How to find existence of such $B$, if possible?
Ref. Chern, Complex Manifolds without Potential Theory pg 84, Sec 9 right around equation (9.3)
Summarizing the comments:
I think the question as stated is false, e.g. if $M$ is just a curve in $\mathbb{P}^2$ then it doesn't miss any lines. But I think that the quotation you are copying is just saying that we can choose $B$ such that $f(M)$ is not entirely contained in $B$, not saying that we can choose $B$ such that $f(M)$ misses $B$.