Is $L = \{ W_1W_2 \mid W_1,W_2 \in (a+b)^* , N_a(W_1) = N_b(W_2)\}$ context free? Can we construct an NPDA for the language?
There is a book here that claims $L$ is not CF (without any elaboration), but I think we can construct a NPDA that accepts the language. My guess is we can construct the language with an NPDA where after reading some $a$ and $b$ and pushing $A$ for each $a$ into the stack, makes a guess to jump to a new state and consumes the pushed $A$ with each $b$.
Yes, the language is context-free.
Let $w$ be a word in $\{a,b\}^*$. Decompose it as $w = u v$ where $|u| = N_b(w)$. $$ \begin{align} N_a(u) &= N_a(w) - N_a(v) && \text{because \(w = u v\)} \\ &= N_a(w) - (|v| - N_b(v)) && \text{because \(v \in \{a,b\}^*\)} \\ &= N_a(w) - |w| + |u| + N_b(v) && \text{because \(|v| = |w| - |u|\)} \\ &= |u| - N_b(w) + N_b(v) = N_b(v) \\ \end{align} $$ Thus we can decompose any word in the desired form: $L = \{a,b\}^*$. Constructing a PDA is left as an exercise to the reader.
This still works if the language contains other letters, by the way, but you have to take a prefix of $w$ that contains $N_b(w)$ letters in $\{a,b\}$ and arbitrarily many other letters.
Another way to see this is to notice that if there is an $a$ in $u$ and a $b$ in $v$, then you can swap the two letters, which results in a word that has the same balance $N_a(u) - N_b(v)$. $L$ is the set of words whose balance is zero for some split position. Iterate this decomposition until it becomes impossible: either $u$ contains no $a$ or $v$ contains no $b$. If we choose the split position so that $|u| = N_b(w)$ and $|v| = N_a(w)$, then now $u$ contains all the $b$'s and $v$ contains all the $a$'s: $N_a(u) = 0$ and $N_b(v) = 0$, and in particular the balance is 0.
