The orbital having one radial node as well one angular node is:

Option A: 3p

$$n=3,\qquad l=1$$

Angular nodes: $$l=1$$

Radial nodes: $$n-l-1=3-1-1=1$$

Thus, 3p has 1 angular node and 1 radial node, matching the question.

Option B: 4f

$$n=4,\qquad l=3$$

Angular nodes: $$l=3$$

Radial nodes: $$n-l-1=4-3-1=0$$

Thus, 4f has 3 angular nodes and 0 radial nodes.

Option C: 4d

$$n=4,\qquad l=2$$

Angular nodes: $$l=2$$

Radial nodes: $$n-l-1=4-2-1=1$$

Thus, 4d has 2 angular nodes and 1 radial node.

Option D: 5d

$$n=5,\qquad l=2$$

Angular nodes: $$l=2$$

Radial nodes: $$n-l-1=5-2-1=2$$

Thus, 5d has 2 angular nodes and 2 radial nodes.

Correct option: (A)