In strict binary tree each node can have either 0 children or 2 children
If no. of leaves are $L$ then it contains $2L-1$ nodes.
Suppose $N$ is total no of nodes, $I$ is total internal nodes and $L$ no of leaves in tree
$N=2*I+1$.....(1)
Also $N= I+L$.....(2)
From (1) and (2)
$I+L=2I+1$
$I=L-1$
$N=2(L-1)+1=2L-1$
With $10$ leaves it contains $2*10-1= 19$ nodes
Hence option B) is correct