Draw functionality unclear for state preparation

Required prerequisites

• Consult the security policy. If reporting a security vulnerability, do not report the bug using this form. Use the process described in the policy to report the issue.
• Make sure you've read the documentation. Your issue may be addressed there.
• Search the issue tracker to verify that this hasn't already been reported. +1 or comment there if it has.
• If possible, make a PR with a failing test to give us a starting point to work on!

Describe the bug

kernel_2 below is a 2q kernel, sample returns a 2q result however draw plots a 1q circuit.

import cudaq 

@cudaq.kernel 
def kernel_1():
    q = cudaq.qvector(1)
    x(q)

state = cudaq.get_state(kernel_1)


@cudaq.kernel
def kernel_2(state: cudaq.State):
    
    a = cudaq.qvector(1)
    b = cudaq.qvector(state)
    
    x(a)

print(cudaq.draw(kernel_2, state))
print(cudaq.sample(kernel_2, state))

     ╭───╮
q0 : ┤ x ├
     ╰───╯

{ 11:1000 }

Steps to reproduce the bug

NA

Expected behavior

NA

Is this a regression? If it is, put the last known working version (or commit) here.

Not a regression

Environment

• CUDA-Q version:
• Python version:
• C++ compiler:
• Operating system:

Suggestions

No response

Start by reproducing the provided Python snippet using cudaq.draw(kernel_2, state) and cudaq.sample(kernel_2, state). Trace the draw entry point and its handling of a cudaq.State, then compare the qubit count with the sample result. Done means the issue's state-preparation behavior is understood and the drawn circuit correctly reflects the state input, or the limitation is documented.