We need the $440^{\text{th}}$ word in the dictionary arrangement of all permutations of the letters of

$\text{KANPUR}$

The letters are all distinct:

$A, K, N, P, R, U$

Arranged alphabetically:

$A < K < N < P < R < U$

Total number of words:

$6! = 720$

We now locate the $440^{\text{th}}$ word step by step.

1. Find the first letter

For each fixed first letter, number of arrangements of remaining 5 letters is:

$5! = 120$

So the blocks are:

• Positions $1$ to $120$: starts with $A$
• Positions $121$ to $240$: starts with $K$
• Positions $241$ to $360$: starts with $N$
• Positions $361$ to $480$: starts with $P$
• Positions $481$ to $600$: starts with $R$
• Positions $601$ to $720$: starts with $U$

Since $440$ lies in $361$ to $480$, the first letter is:

$P$

Relative position within the $P$-block:

$440 - 360 = 80$

So now we need the $80^{\text{th}}$ arrangement of the remaining letters:

$A, K, N, R, U$

2. Find the second letter

For each fixed second letter, arrangements of remaining 4 letters:

$4! = 24$

Alphabetical order of remaining letters:

$A, K, N, R, U$

Blocks inside the $P$-block:

• $PA$: positions $1$ to $24$
• $PK$: positions $25$ to $48$
• $PN$: positions $49$ to $72$
• $PR$: positions $73$ to $96$
• $PU$: positions $97$ to $120$

The relative position is $80$, which lies in $73$ to $96$. So second letter is:

$R$

New relative position within the $PR$-block:

$80 - 72 = 8$

Now we need the $8^{\text{th}}$ arrangement of:

$A, K, N, U$

3. Find the third letter

For each fixed third letter, arrangements of remaining 3 letters:

$3! = 6$

Alphabetical order:

$A, K, N, U$

Blocks:

• $PRA$: positions $1$ to $6$
• $PRK$: positions $7$ to $12$
• $PRN$: positions $13$ to $18$
• $PRU$: positions $19$ to $24$

The relative position is $8$, which lies in $7$ to $12$. So third letter is:

$K$

New relative position within the $PRK$-block:

$8 - 6 = 2$

Now we need the $2^{\text{nd}}$ arrangement of:

$A, N, U$

4. Find the fourth letter

For each fixed fourth letter, arrangements of remaining 2 letters:

$2! = 2$

Alphabetical order:

$A, N, U$

Blocks:

• $PRKA$: positions $1$ to $2$
• $PRKN$: positions $3$ to $4$
• $PRKU$: positions $5$ to $6$

The relative position is $2$, so it lies in the first block. Thus fourth letter is:

$A$

New relative position within $PRKA$:

$2$

Now arrange remaining letters:

$N, U$

Their dictionary order is:

1. $NU$
2. $UN$

The $2^{\text{nd}}$ arrangement is:

$UN$

So the full word is:

$\boxed{PRKAUN}$

5. Compare with options

• A: $PRNAKU$
• B: $PRKAUN$
• C: $PRKANU$
• D: $PRNAUK$

Hence the correct option is:

$\boxed{\text{B}}$

6. Comparison with stored correct answer

Stored correct answer: $\text{B}$

Our derived answer is also $\text{B}$, so they agree.