Let $\mathbb{Z^{+}}=\mathbb{N}\cup\{0\}=\{0,1,2,3,4,5,...\}$
$\begin{align}\therefore \mathbb{Z^{+}}\times \mathbb{Z^{+}}=\{&(0,0),(0,1),(0,2),...,\\&(1,0),(1,1),(1,2),...,\\&(2,0),(2,1),(2,2),...,\\&.....................,\\&(100,0),(100,1),(100,2),(100,3),...,\\&.....................\}\end{align}$
To answer this question, we need to think of the worst case. How many pairs are there at least to take from $\mathbb{Z^{+}}\times \mathbb{Z^{+}}$ where for any two pairs $(a,b),(c,d)$ the condition $a \equiv c \mod 3$ and $b \equiv d \mod 5 $ doesn't hold?
The answer is $3\times5=15$.
Proof: For any integer $x \mathrm{~mod~}3$ has the residues from the set $\{0,1,2\}=A$ [Let]. Likewise for $x \mathrm{~mod~} 5$ has the residues from the set $\{0,1,2,3,4\}=B$ [Let].
Now,
$\begin{align} A \times B=\{&(0,0),(0,1),(0,2),(0,3),(0,4),\\&(1,0),(1,1),(1,2),(1,3),(1,4),\\&(2,0),(2,1),(2,2),(2,3),(2,4)\}\end{align}$
This set of ordered pairs doesn't hold the condition because there are no two pairs $(a,b),(c,d)$ such that $a \equiv c \mod 3$ and $b \equiv d \mod 5 $.
Here, $|A \times B|=|A|\times|B|=3 \times 5=15$.
So for the worst case, we have to take at least $15$ pairs from $\mathbb{Z^{+}}\times \mathbb{Z^{+}}$. Now taking any other pair (only one more) will hold the required condition. Because for any other pair $(x,y)$, we have $(x\mod 3, y\mod5)$ will produce any pair from $A \times B$. [For example, take $(95,101)$, then we can have $(95\mod3,101\mod5)\equiv(2,1)\in A \times B$].
Therefore, we have to take $15+1=16$ pairs from $\mathbb{Z^{+}}\times \mathbb{Z^{+}}$.
So the correct answer is C.