B is 100% correct.
D is the closest answer than others. hosts connected to the same switch DOES NOT mean that they are on the same subnet. Different switch ports can be on different subnets (This is what VLANs are for, breaking down broadcast domains in a switch)
as others already mentioned, a /24 network can only have 254 host addresses
Agreed. PC-B could easily be sitting in another VLAN. Option D is vague to the point of absurdity but it's the only option that will satisfy the question( Cisco should know better!) B&D are the best choices.
erratum
B & D are correct,
E is wrong since there can only be 254 hosts on the subnet of PC C since 2 ip addresses must be assigned to network and broadcast address
1 is network address and the other is the broadcast address, these two are not Hosts. So, only Hosts or usable hosts( if you want to call them like this) : 254. Doesn't exist unusable hosts.
The thing here is, PC-B and PC-A could be on the same subnet (192.168.3.0/24) OR that PC-A and PC-B are on the same subnet (192.168.1.0/24), this is not clear on the picture. Now if we use an IP calculator on PC-C network address, we have 2 raw information, Total Host (256) and total usable host (254). the answer E says 256 hosts, not usable hosts, so E could be right answer, depending on what I mention at the beginning.
B & D. the CID notation /24 or 255.255.255.0 leave 8 bit for the host and 2 ^ 8 is 256 but you have to minus 2 which reduces to 254 host; The first is the network and the last is the broadcast. so you can not have 256 as the answer
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